In my opinion, I feel like this is definitely not achieving what it's supposed to do. I was needing a random number generator based off probability to determine a winner of a race. So Runner A has a 40% chance of winning, for example.
//function definition
void createDogs(DOGS* dogList) {
//adding info the the dogInfo struct
strcpy(dogList[0].dogName, "Easy Rex"); //dog 1
dogList[0].odds = 40;
dogList[0].payoutMultiplier = 2;
strcpy(dogList[1].dogName, "Worried Bud"); //dog 2
dogList[1].odds = 10;
dogList[1].payoutMultiplier = 5;
strcpy(dogList[2].dogName, "Money Ace"); //dog 3
dogList[2].odds = 8;
dogList[2].payoutMultiplier = 10;
strcpy(dogList[3].dogName, "Lucky Lady"); //dog 4
dogList[3].odds = 15;
dogList[3].payoutMultiplier = 15;
strcpy(dogList[4].dogName, "Cash Dawg"); //dog 5
dogList[4].odds = 1;
dogList[4].payoutMultiplier = 50;
strcpy(dogList[5].dogName, "Unlucky Brutus"); //dog 6
dogList[5].odds = 4;
dogList[5].payoutMultiplier = 20;
strcpy(dogList[6].dogName, "Gamble Champ"); //dog 7
dogList[6].odds = 8;
dogList[6].payoutMultiplier = 10;
strcpy(dogList[7].dogName, "Nothing Chewy"); //dog 8
dogList[7].odds = 10;
dogList[7].payoutMultiplier = 5;
strcpy(dogList[8].dogName, "Easy Roxy"); //dog 9
dogList[8].odds = 13;
dogList[8].payoutMultiplier = 3;
}//end createDogs
So here's where I put the probability, under the "odds", then here is where I implement it. Thinking that the odds is the percentage so grabbing the random number from 0 to that "percentage" and then compare them with the others to determine the winner.
//function definition
void dogRace(DOGS* dogList, DATA* raceInfo, int counter) {
int numberRolled[NO_OF_DOGS];
int i, moneyWon;
int biggestNumber, position = 0;
srand(time(0));
printf("\nAnd the race is on!");
pause(5);
for (i = 0; i < NO_OF_DOGS; i++) { //assigns a number to each dog based on its odds
numberRolled[i] = (rand() % dogList[i].odds);
}//end for
biggestNumber = numberRolled[0];
for (i = 0; i < NO_OF_DOGS; i++) { //determines which dog won (>number rolled)
if (biggestNumber < numberRolled[i]) {
biggestNumber = numberRolled[i];
position = i;
}//end if
}//end for
}
Technically it works, but I feel as it's not really the right way to do it- let me know if there is a better way to do this because my textbooks do no give any example for this type of problem. I've also tried googling/youtubing it and had no luck finding for what I was looking for.
Try using cumulative odds:
S = sum of odds of dogs 0 to number of dogs - 1
R = random integer from 0 to S-1
i = 0
while i < number of dogs and dog[i].odds <= R: R = R - dog[i].odds, i = i + 1
//function definition
void dogRace(DOGS* dogList, DATA* raceInfo, int counter) {
int i, moneyWon, position;
int sum_of_odds = 0;
int rolled;
for (i = 0; i < NO_OF_DOGS; i++) {
sum_of_odds += dogList[i].odds;
}
// Note, usually you should only call srand() once in a program.
srand(time(0));
printf("\nAnd the race is on!");
pause(5);
rolled = randInt(sum_of_odds);
for (i = 0; i < NO_OF_DOGS; i++) {
if (dogList[i].odds > rolled)
break;
rolled -= dogList[i].odds;
}//end for
position = i;
}
The randInt function called above is given below:
// random integer from 0 to n-1 (for n in range 1 to RAND_MAX+1u)
int randInt(unsigned int n) {
unsigned int x = (RAND_MAX + 1u) / n;
unsigned int limit = x * n;
int s;
do {
s = rand();
} while (s >= limit);
return s / x;
}
The above is preferable to using rand() % n because it removes any bias in the likely case that RAND_MAX+1 is not a multiple of n. Also, some implementations of rand() produce not very random sequences for rand() % n so it is better to use the quotient of division rather than the remainder.
I am fighting some simple question.
I want to get prime numbers
I will use this algorithm
and... I finished code writing like this.
int k = 0, x = 1, n, prim, lim = 1;
int p[100000];
int xCount=0, limCount=0, kCount=0;
p[0] = 2;
scanf("%d", &n);
start = clock();
do
{
x += 2; xCount++;
if (sqrt(p[lim]) <= x)
{
lim++; limCount++;
}
k = 2; prim = true;
while (prim && k<lim)
{
if (x % p[k] == 0)
prim = false;
k++; kCount++;
}
if (prim == true)
{
p[lim] = x;
printf("prime number : %d\n", p[lim]);
}
} while (k<n);
I want to check how much repeat this code (x+=2; lim++; k++;)
so I used xCount, limCount, kCount variables.
when input(n) is 10, the results are x : 14, lim : 9, k : 43. wrong answer.
answer is (14,3,13).
Did I write code not well?
tell me correct point plz...
If you want to adapt an algorithm to your needs, it's always a good idea to implement it verbatim first, especially if you have pseudocode that is detailed enough to allow for such a verbatim translation into C-code (even more so with Fortran but I digress)
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
int main (void){
// type index 1..n
int index;
// var
// x: integer
int x;
//i, k, lim: integer
int i, k, lim;
// prim: boolean
bool prim;
// p: array[index] of integer {p[i] = i'th prime number}
/*
We cannot do that directly, we need to know the value of "index" first
*/
int res;
res = scanf("%d", &index);
if(res != 1 || index < 1){
fprintf(stderr,"Only integral values >= 1, please. Thank you.\n");
return EXIT_FAILURE;
}
/*
The array from the pseudocode is a one-based array, take care
*/
int p[index + 1];
// initialize the whole array with distinguishable values in case of debugging
for(i = 0;i<index;i++){
p[i] = -i;
}
/*
Your variables
*/
int lim_count = 0, k_count = 0;
// begin
// p[1] = 2
p[1] = 2;
// write(2)
puts("2");
// x = 1
x = 1;
// lim = 1
lim = 1;
// for i:=2 to n do
for(i = 2;i < index; i++){
// repeat (until prim)
do {
// x = x + 2
x += 2;
// if(sqr(p[lim]) <= x) then
if(p[lim] * p[lim] <= x){
// lim = lim +1
lim++;
lim_count++;
}
// k = 2
k = 2;
// prim = true
prim = true;
// while (prim and (k < lim)) do
while (prim && (k < lim)){
// prim = "x is not divisible by p[k]"
if((x % p[k]) == 0){
prim = false;
}
// k = k + 1
k++;
k_count++;
}
// (repeat) until prim
} while(!prim);
// p[i] := x
p[i] = x;
// write(x)
printf("%d\n",x);
}
// end
printf("x = %d, lim_count = %d, k_count = %d \n",x,lim_count,k_count);
for(i = 0;i<index;i++){
printf("%d, ",p[i]);
}
putchar('\n');
return EXIT_SUCCESS;
}
It will print an index - 1 number of primes starting at 2.
You can easily change it now--for example: print only the primes up to index instead of index - 1 primes.
In your case the numbers for all six primes up to 13 gives
x = 13, lim_count = 2, k_count = 3
which is distinctly different from the result you want.
Your translation looks very sloppy.
for i:= 2 to n do begin
must translate to:
for (i=2; i<=n; i++)
repeat
....
until prim
must translate to:
do {
...
} while (!prim);
The while prim... loop is inside the repeat...until prim loop.
I leave it to you to apply this to your code and to check that all constructs have been properly translated. it doesn't look too difficult to do that correctly.
Note: it looks like the algorithm uses 1-based arrays whereas C uses 0-based arrays.
The problem is the following: Given "ABC+DEF=GHI" format string, where A,B,C etc. represent unique digits, find the expression that gives maximum GHI. Ex: Input string is AAB+AAB=AAB, then there's no solution. If it is instead AAA + BBB = AAA, a solution is 999 + 000 = 999. Another example string: ABC + CBA = GGG, a result is => 543 + 345 = 888.
I have ruled out impossible cases easily. The algorithm I have in mind is a bruteforce, that simply tries maximizing the rhs first. However my problem was doing this fast, and also watching out for the unique digits. What's an efficient way to solve this problem?
Notes: I wish to solve this in a singlethreaded approach, and my current problem is detecting if a unique digit is used in "assign_value" function. Perhaps a better method to assign values is there?
EDIT: As per smci's suggestion, here's what I want to achieve, in the very end: ABRA + CADABRA + ABRA + CADABRA == HOUDINI ; 7457 + 1797457 + 7457 + 1797457 == 3609828 -- A system that can handle not only strings of the form I provided in the beginning (3 digit number + 3 digit number = 3 digit number) but also those. However it doesn't hurt to start simple and go with the solution of format I gave :)
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_EXPRESSION_SIZE 11 + 1
#define MAX_VARIABLES 9
int variables_read[MAX_VARIABLES] = { 0 };
struct variable {
int coefficient;
int* ptr;
int side;
int canhavezero;
unsigned value_max;
};
typedef struct variable Variable;
struct equation {
Variable* variables[9]; // max
unsigned distinct_on_rhs;
unsigned var_count;
};
typedef struct equation Equation;
int int_pow(int n, int k) {
int res = 1;
for(int i = 0; i < k; ++i)
res *= n;
return res;
}
void AddVariable(Equation* E, Variable* V) {
E->variables[E->var_count++] = V;
}
int IsImpossible(char* expression) {
// if all letters are same or end letters are same, no solution
if(
(expression[0] == expression[4] && expression[0] == expression[8]) ||
(!strncmp(expression, expression + 4, 3) && !strncmp(expression, expression + 8, 3))
)
return 1;
return 0;
}
int assign_value(Equation* E, int pos, int* values) {
if(!E->variables[pos]->value_count) {
if(pos < 0)
return 2;
// if no possible values left, reset this, but take one value count from the closest variable
E->variables[pos - 1]->value_count--;
E->variables[pos]->value_count = E->variables[pos]->value_max;
return 0;
}
int i;
for(i = 9; i >= 0 && values[i] == -1; --i)
printf("Assigning %d to %c\n", E->variables[pos]->value_set[E->variables[pos]->value_count - 1], 'A' + (E->variables[pos]->ptr - E->variables[0]->ptr));
*(E->variables[pos]->ptr) = values[i];
values[i] = -1; // we have unique numbers
return 0;
}
int isSolved(Equation E) {
int sum = 0, coeff = 0;
printf("Trying...\n");
for(int i = 0; i < E.var_count; ++i) {
coeff = E.variables[i]->coefficient * (*E.variables[i]->ptr);
printf("%d ", *E.variables[i]->ptr);
if(E.variables[i]->side)
coeff *= -1;
sum += coeff;
}
printf("\nSum was %d\n", sum);
return !sum;
}
char* evaluate(char* expression) {
char* res;
// check for impossible cases first
if(IsImpossible(expression)) {
res = (char *) malloc(sizeof(char) * strlen("No Solution!"));
strcpy(res, "No Solution!");
return res;
}
res = (char *) malloc(sizeof(char) * MAX_EXPRESSION_SIZE);
// now try to find solutions, first describe the given characters as equations
Equation E;
E.var_count = 0;
E.distinct_on_rhs = 0;
int side_mode = 0, powcounter = 0;
int a = -1, b = -1, c = -1, d = -1, e = -1, f = -1, g = -1, h = -1, i = -1;
int* max_variables[MAX_VARIABLES] = { &a, &b, &c, &d, &e, &f, &g, &h, &i };
for(int j = 0; j < MAX_EXPRESSION_SIZE - 1; ++j) {
if(expression[j] == '+')
continue;
if(expression[j] == '=') {
side_mode = 1;
continue;
}
Variable* V = (Variable *) malloc(sizeof(Variable));
// we know we always get 3 digit numbers but we can easily change if we need to
V->coefficient = int_pow(10, 2 - (powcounter % 3));
V->ptr = max_variables[expression[j] - 'A'];
V->side = side_mode;
E.distinct_on_rhs += side_mode && !variables_read[expression[j] - 'A'];
if(!(powcounter % 3)) { // beginning of a number
V->value_count = 9;
V->value_max = 9;
V->canhavezero = 0;
}
else {
V->value_count = 10;
V->value_max = 10;
V->canhavezero = 1;
}
AddVariable(&E, V);
variables_read[expression[j] - 'A'] = 1;
++powcounter;
}
for(int j = 0; j < E.var_count; ++j)
printf("%d %c %d\n", E.variables[j]->coefficient, 'A' + (E.variables[j]->ptr - max_variables[0]), E.variables[j]->side);
// we got a representaion of the equation, now try to solve it
int solved = 0;
// O(9^N), where N is number of distinct variables.
// An optimization we can do is, we first assign possible max values to rhs number, then go down. We need max number.
printf("Distincts: %d\n", E.distinct_on_rhs);
do {
// try to assign values to all variables and try if it solves the equation
// but first try to assign rhs as max as possible
int values[10] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
int temp = E.var_count - E.distinct_on_rhs;
while(temp < E.var_count) {
solved = assign_value(&E, temp, values);
++temp;
}
for(int j = E.var_count - 1 - E.distinct_on_rhs; j >= 0; --j)
solved = assign_value(&E, j, values);
if(solved) // can return no solution
break;
printf("Solving...\n");
solved = isSolved(E);
system("PAUSE");
} while(!solved);
if(solved == 2) {
res = (char *) malloc(sizeof(char) * strlen("No Solution!"));
strcpy(res, "No Solution!");
}
else {
}
return res;
}
int main() {
char expression[MAX_EXPRESSION_SIZE] = { 0 };
do {
printf("Enter the formula: ");
scanf("%s", expression);
char* res = evaluate(expression);
printf("%s\n", res);
free(res);
} while(expression[0] != '-');
return 0;
}
I would start with the result. There are not that many different cases:
AAA
AAB, ABA, BAA
ABC
All other cases can be reduced to these by renaming the variables. ABC + CBA = GGG would become DBC + CBD = AAA.
Then you have
10 possible solutions for the one-variable case AAA
90 (10*9) for the two variable cases
720 (10*9*8) for the three variable case
assuming that zero is allowed anywhere. If not, you can filter out those that are not allowed.
This sets the variables for the right side of the equation. Each variable that appears only on the left, adds possible solutions. B adds a factor of 9, C a factor of 8, D 7 and so forth.
The most "efficient" solution would take all knowledge of the task and simple print the result. So the question is how much of the conditions can be coded and where and what flexibility is needed.
An alternative is to view the generation of test cases and evaluation of them separately.
A simple recursion function can generate the 10! (362880) test cases of unique digits.
unsigned long long count = 0;
unsigned long long sol = 0;
void evaluate(int object[]) {
count++;
int ABC = object[0] * 100 + object[1] * 10 + object[2];
int DEF = object[3] * 100 + object[4] * 10 + object[5];
int GHI = object[6] * 100 + object[7] * 10 + object[8];
if (ABC + DEF == GHI) {
printf("%4llu %03d + %03d = %03d\n", ++sol, ABC,DEF,GHI);
}
}
void form_combos(int pool[], size_t pool_count, int object[],
size_t object_count, size_t object_count_max) {
if (object_count >= object_count_max) {
evaluate(object);
return;
}
assert(pool_count > 0);
int *pool_end = pool + pool_count - 1;
for (size_t p = 0; p < pool_count; p++) {
int sample = pool[p]; // take one out
pool[p] = *pool_end; // replace it with the end
object[object_count] = sample;
form_combos(pool, pool_count - 1, object, object_count + 1,
object_count_max);
pool[p] = sample; // restore pool item
}
}
int main() {
int pool[] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
size_t pool_size = sizeof pool / sizeof pool[0];
#define object_count 9
int object[object_count];
form_combos(pool, pool_size, object, 0, object_count);
printf("Evaluate() iterations %llu\n", count);
}
Output
1 091 + 762 = 853
2 091 + 763 = 854
3 091 + 735 = 826
...
1726 874 + 061 = 935
1727 875 + 046 = 921
1728 876 + 045 = 921
Evaluate() iterations 3628800
What is nice about this approach is that if the task was now find
ABC*ABC + DEF*DEF == GHI*GHI
Changing only 2 lines of code:
if (ABC*ABC + DEF*DEF == GHI*GHI) {
printf("%4llu sqr(%03d) + sqr(%03d) = sqr(%03d)\n", ++sol, ABC,DEF,GHI);
}
results in
1 sqr(534) + sqr(712) = sqr(890)
2 sqr(546) + sqr(728) = sqr(910)
3 sqr(712) + sqr(534) = sqr(890)
4 sqr(728) + sqr(546) = sqr(910)
Evaluate() iterations 3628800
Ok, so for a trivial solution (a base to build a generalization on, so far it only works on the format <3 digit number> + <3 digit number> = <3 digit number>) inspired from #chux and #alain's suggestions is the following code. It truly runs on O(10^N) where N is the distinct number of digits present, or variables if you'd like to call them that. I'll see if I can generalize this even further.
Note that this is for the initial problem of finding the largest rhs. Take that into account as well.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_DIGITS 10
#define MAX_VARIABLES 9
#define MAX_EXPRESSION_SIZE 11
int IsImpossible(char* expression) {
// if all letters are same or end letters are same, no solution
if(
(expression[0] == expression[4] && expression[0] == expression[8]) ||
(!strncmp(expression, expression + 4, 3) && !strncmp(expression, expression + 8, 3))
)
return 1;
return 0;
}
int ArePointersAssigned(int*** pointers) {
for(int i = 0; i < MAX_VARIABLES; ++i) {
if(**pointers[i] == -1)
return 0;
}
return 1;
}
int evaluate(int*** pointers) {
int ABC = *(*pointers[0]) * 100 + *(*pointers[1]) * 10 + *(*pointers[2]);
int DEF = *(*pointers[3]) * 100 + *(*pointers[4]) * 10 + *(*pointers[5]);
int GHI = *(*pointers[6]) * 100 + *(*pointers[7]) * 10 + *(*pointers[8]);
if (ABC + DEF == GHI) { // since we use dfs, if this is a solution simply return it
//printf("%d + %d = %d\n", ABC, DEF, GHI);
return 1;
}
return 0;
}
// use the solved pointer to escape recursion early
// check_end checks if we reached 6 for the 2nd time, if it's first time we ignore (because it's start state)
void form_combos(int pool[], int pool_count, int object_count, int*** pointers, int* solved) {
if(object_count == MAX_DIGITS - 1)
object_count = 0;
if(*solved) // if a branch solved this, escape recursion
return;
if (ArePointersAssigned(pointers)) { // that means we got a full equation set
*solved = evaluate(pointers);
if(*solved)
return;
}
int *pool_end = pool + pool_count - 1;
for (int p = pool_count - 1; p >= 0 && !*solved; p--) {
int sample = pool[p]; // take one out
pool[p] = *pool_end; // replace it with the end
int temp = **pointers[object_count];
if(**pointers[object_count] == -1)
**pointers[object_count] = sample;
form_combos(pool, pool_count - 1, object_count + 1, pointers, solved);
pool[p] = sample; // restore pool item
if(!*solved)
**pointers[object_count] = temp;
}
}
int main() {
char expression[MAX_EXPRESSION_SIZE] = { 0 };
printf("Enter the formula: ");
scanf("%s", expression);
while(expression[0] != '-') {
if(IsImpossible(expression))
printf("No solution!\n");
else {
int digits[MAX_DIGITS] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
int object[MAX_VARIABLES] = { -1, -1, -1, -1, -1, -1, -1, -1, -1 }; // stack for dfs
int *A = &object[0], *B = &object[1], *C = &object[2],
*D = &object[3], *E = &object[4], *F = &object[5],
*G = &object[6], *H = &object[7], *I = &object[8];
// set same pointers
int** pointers[MAX_VARIABLES] = { &A, &B, &C, &D, &E, &F, &G, &H, &I };
// analyze the equation
int var = 0;
for(int p = 0; p < MAX_EXPRESSION_SIZE; ++p) {
if(expression[p] >= 'A' && expression[p] <= 'I') {
*pointers[var++] = &object[expression[p] - 'A']; // link same pointers
}
}
int solved = 0, check_end = 0;
form_combos(digits, MAX_DIGITS, MAX_DIGITS - 4, pointers, &solved);
if(!solved) // it can be unsolvable still
printf("No solution!\n");
else
printf("%d%d%d + %d%d%d = %d%d%d\n", *A, *B, *C, *D, *E, *F, *G, *H, *I);
}
printf("Enter the formula: ");
scanf("%s", expression);
}
return 0;
}
So I need to fill in the code for the program to work for the question:
Array testGrades contains NUM_VALS test scores. Write a for loop that sets sumExtra to the total extra credit received. Full credit is 100, so anything over 100 is extra credit. Ex: If testGrades = {101, 83, 107, 90}, then sumExtra = 8, because 1 + 0 + 7 + 0 is 8.
#include <stdio.h>
int main(void) {
const int NUM_VALS = 4;
int testGrades[NUM_VALS];
int i = 0;
int sumExtra = -9999; // Initialize to 0 before your for loop
testGrades[0] = 101;
testGrades[1] = 83;
testGrades[2] = 107;
testGrades[3] = 90;
// STUDENT CODE GOES HERE
return 0;
}
So far all I have is:
for (i=0;i<NUM_VALS;++i) {
if (testGrades[i] > 100) {
sumExtra = testGrades[i] - 100;
}
}
I dont know how to find the sum of the array of the values over 100.
Firstly, initialise sumExtra to 0. Then just change:
sumExtra = testGrades[i] - 100;
to
sumExtra += testGrades[i] - 100;
because the sumExtra for a particular index i is testGrades[i]-100, and you want to find the total of the sumExtra, and hence keeping adding this to the sumExtra variable.
You are missing the if statement that checks if the test score is above 100. Here is the code that works:
sumExtra = 0;
for (i = 0; i < NUM_VALS; ++i) {
if (testGrades[i] >= 101) {
sumExtra += (testGrades[i] - 100);
}
}
cout << "sumExtra: " << sumExtra << endl;
Initialize to 0 before your for loop
int sumExtra = 0;
sumExtra = testGrades[i]-100;
sumExtra += testGrades[i]-100;
For solving project euler problem 20 to find the sum of digits in 100! i am running the following program , it is working for factorial of small numbers but not for 100.which data type should i use or is it necessary to use an array for storing the digits?
int rec(int);
void main()
{
int f=1,i=1,z,s=0,r,n;
while(i<=100)
{
f=f*i;
f=rec(f);
i++;
}
n=f;
while(n!=0)
{
r=n%10;
n=n/10;
s=s+r;
}
printf("\n%d",s);
}
int rec(int t)
{
if(t%10==0)
{
t=t/10;
rec(t);
}
return t;
}
Approximate factorial of 100 can be calculated using the double type. You can also use the Stirling's formula, stating that
n! ≈ sqrt(2*M_PI*n) * pow(n/exp(0),n)
If you plug in the numbers, you'll get n! ≈ 9*10157. That means your type needs to be able to hold 158 decimal digits or, equivalently, ~log2(9*10157) = 525 bits or 66 8-bit bytes.
No fundamental numeric type in C is big enough. The largest you are guaranteed to get is 64 bits (if you use unsigned long long).
So, if you want to calculate n! in C, you either need to construct long arithmetic multiplication by hand or use a special library that can do that for you.
For this relatively simple task you can actually implement long multiplication and use it to get the factorial value by repeated multiplication.
In the following program I've used an in-place multiplication algorithm, which modifies one of the multiplicands in the process and eventually replaces it with the product. The algorithm can be derived directly from the long multiplication known from school.
This program calculates factorials of integers from 1 up to and including 100:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <limits.h>
typedef unsigned char uint8;
typedef unsigned short uint16;
#if UINT_MAX >= 0xFFFFFFFF
typedef unsigned uint32;
#else
typedef unsigned long uint32;
#endif
typedef unsigned uint;
void MulInPlace(uint8* dst/* n bytes */,
const uint8* src/* n bytes */,
uint n)
{
uint c1, c2;
if (n >= 0xFFFF) abort();
for (c1 = n - 1; c1 != ~0u; c1--)
{
uint16 s = 0;
uint32 p = 0; // p must be able to store ceil(log2(n))+2*8 bits
for (c2 = c1; c2 != ~0u; c2--)
{
p += dst[c2] * src[c1 - c2];
}
dst[c1] = (uint8)(p & 0xFF);
for (c2 = c1 + 1; c2 < n; c2++)
{
p >>= 8;
s += dst[c2] + (uint8)(p & 0xFF);
dst[c2] = (uint8)(s & 0xFF);
s >>= 8;
}
}
}
int ByteDivInPlace(uint8* dst/* n bytes */,
uint n,
uint8 divisor,
uint8* remainder)
{
uint rem = 0;
int nonzero = 0;
while (n)
{
rem += dst[n - 1];
nonzero |= (dst[n - 1] = rem / divisor);
rem = (rem % divisor) << 8;
n--;
}
if (remainder != NULL)
*remainder = (uint8)(rem >> 8);
return nonzero; // 1 if the quotient is non-zero, 0 otherwise
}
void IncInPlace(uint8* dst/* n bytes */,
uint n)
{
uint c = 1;
while (n-- && c)
{
c += *dst;
*dst++ = c & 0xFF;
c >>= 8;
}
}
void DestroyingDecimalPrint(uint8* dst, uint n)
{
uint8 r;
if (ByteDivInPlace(dst, n, 10, &r))
DestroyingDecimalPrint(dst, n);
printf("%d", r);
}
int main(void)
{
int i;
uint8 factorial[66];
uint8 factor[sizeof(factorial)];
uint8 tmp[sizeof(factorial)];
// factor = 1
memset(factor, 0, sizeof(factor));
factor[0] = 1;
// factorial = 1
memcpy(factorial, factor, sizeof(factorial));
for (i = 1; i <= 100; i++)
{
// factorial *= factor
MulInPlace(factorial, factor, sizeof(factorial));
// tmp = factorial
memcpy(tmp, factorial, sizeof(factorial));
// print i and tmp
printf("%i! = ", i);
DestroyingDecimalPrint(tmp, sizeof(tmp));
printf("\n");
// factor += 1
IncInPlace(factor, sizeof(factor));
}
return 0;
}
Output (ideone):
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = 5040
8! = 40320
9! = 362880
10! = 3628800
11! = 39916800
12! = 479001600
13! = 6227020800
14! = 87178291200
15! = 1307674368000
16! = 20922789888000
17! = 355687428096000
18! = 6402373705728000
19! = 121645100408832000
20! = 2432902008176640000
21! = 51090942171709440000
22! = 1124000727777607680000
23! = 25852016738884976640000
24! = 620448401733239439360000
25! = 15511210043330985984000000
26! = 403291461126605635584000000
27! = 10888869450418352160768000000
28! = 304888344611713860501504000000
29! = 8841761993739701954543616000000
30! = 265252859812191058636308480000000
31! = 8222838654177922817725562880000000
32! = 263130836933693530167218012160000000
33! = 8683317618811886495518194401280000000
34! = 295232799039604140847618609643520000000
35! = 10333147966386144929666651337523200000000
36! = 371993326789901217467999448150835200000000
37! = 13763753091226345046315979581580902400000000
38! = 523022617466601111760007224100074291200000000
39! = 20397882081197443358640281739902897356800000000
40! = 815915283247897734345611269596115894272000000000
41! = 33452526613163807108170062053440751665152000000000
42! = 1405006117752879898543142606244511569936384000000000
43! = 60415263063373835637355132068513997507264512000000000
44! = 2658271574788448768043625811014615890319638528000000000
45! = 119622220865480194561963161495657715064383733760000000000
46! = 5502622159812088949850305428800254892961651752960000000000
47! = 258623241511168180642964355153611979969197632389120000000000
48! = 12413915592536072670862289047373375038521486354677760000000000
49! = 608281864034267560872252163321295376887552831379210240000000000
50! = 30414093201713378043612608166064768844377641568960512000000000000
51! = 1551118753287382280224243016469303211063259720016986112000000000000
52! = 80658175170943878571660636856403766975289505440883277824000000000000
53! = 4274883284060025564298013753389399649690343788366813724672000000000000
54! = 230843697339241380472092742683027581083278564571807941132288000000000000
55! = 12696403353658275925965100847566516959580321051449436762275840000000000000
56! = 710998587804863451854045647463724949736497978881168458687447040000000000000
57! = 40526919504877216755680601905432322134980384796226602145184481280000000000000
58! = 2350561331282878571829474910515074683828862318181142924420699914240000000000000
59! = 138683118545689835737939019720389406345902876772687432540821294940160000000000000
60! = 8320987112741390144276341183223364380754172606361245952449277696409600000000000000
61! = 507580213877224798800856812176625227226004528988036003099405939480985600000000000000
62! = 31469973260387937525653122354950764088012280797258232192163168247821107200000000000000
63! = 1982608315404440064116146708361898137544773690227268628106279599612729753600000000000000
64! = 126886932185884164103433389335161480802865516174545192198801894375214704230400000000000000
65! = 8247650592082470666723170306785496252186258551345437492922123134388955774976000000000000000
66! = 544344939077443064003729240247842752644293064388798874532860126869671081148416000000000000000
67! = 36471110918188685288249859096605464427167635314049524593701628500267962436943872000000000000000
68! = 2480035542436830599600990418569171581047399201355367672371710738018221445712183296000000000000000
69! = 171122452428141311372468338881272839092270544893520369393648040923257279754140647424000000000000000
70! = 11978571669969891796072783721689098736458938142546425857555362864628009582789845319680000000000000000
71! = 850478588567862317521167644239926010288584608120796235886430763388588680378079017697280000000000000000
72! = 61234458376886086861524070385274672740778091784697328983823014963978384987221689274204160000000000000000
73! = 4470115461512684340891257138125051110076800700282905015819080092370422104067183317016903680000000000000000
74! = 330788544151938641225953028221253782145683251820934971170611926835411235700971565459250872320000000000000000
75! = 24809140811395398091946477116594033660926243886570122837795894512655842677572867409443815424000000000000000000
76! = 1885494701666050254987932260861146558230394535379329335672487982961844043495537923117729972224000000000000000000
77! = 145183092028285869634070784086308284983740379224208358846781574688061991349156420080065207861248000000000000000000
78! = 11324281178206297831457521158732046228731749579488251990048962825668835325234200766245086213177344000000000000000000
79! = 894618213078297528685144171539831652069808216779571907213868063227837990693501860533361810841010176000000000000000000
80! = 71569457046263802294811533723186532165584657342365752577109445058227039255480148842668944867280814080000000000000000000
81! = 5797126020747367985879734231578109105412357244731625958745865049716390179693892056256184534249745940480000000000000000000
82! = 475364333701284174842138206989404946643813294067993328617160934076743994734899148613007131808479167119360000000000000000000
83! = 39455239697206586511897471180120610571436503407643446275224357528369751562996629334879591940103770870906880000000000000000000
84! = 3314240134565353266999387579130131288000666286242049487118846032383059131291716864129885722968716753156177920000000000000000000
85! = 281710411438055027694947944226061159480056634330574206405101912752560026159795933451040286452340924018275123200000000000000000000
86! = 24227095383672732381765523203441259715284870552429381750838764496720162249742450276789464634901319465571660595200000000000000000000
87! = 2107757298379527717213600518699389595229783738061356212322972511214654115727593174080683423236414793504734471782400000000000000000000
88! = 185482642257398439114796845645546284380220968949399346684421580986889562184028199319100141244804501828416633516851200000000000000000000
89! = 16507955160908461081216919262453619309839666236496541854913520707833171034378509739399912570787600662729080382999756800000000000000000000
90! = 1485715964481761497309522733620825737885569961284688766942216863704985393094065876545992131370884059645617234469978112000000000000000000000
91! = 135200152767840296255166568759495142147586866476906677791741734597153670771559994765685283954750449427751168336768008192000000000000000000000
92! = 12438414054641307255475324325873553077577991715875414356840239582938137710983519518443046123837041347353107486982656753664000000000000000000000
93! = 1156772507081641574759205162306240436214753229576413535186142281213246807121467315215203289516844845303838996289387078090752000000000000000000000
94! = 108736615665674308027365285256786601004186803580182872307497374434045199869417927630229109214583415458560865651202385340530688000000000000000000000
95! = 10329978488239059262599702099394727095397746340117372869212250571234293987594703124871765375385424468563282236864226607350415360000000000000000000000
96! = 991677934870949689209571401541893801158183648651267795444376054838492222809091499987689476037000748982075094738965754305639874560000000000000000000000
97! = 96192759682482119853328425949563698712343813919172976158104477319333745612481875498805879175589072651261284189679678167647067832320000000000000000000000
98! = 9426890448883247745626185743057242473809693764078951663494238777294707070023223798882976159207729119823605850588608460429412647567360000000000000000000000
99! = 933262154439441526816992388562667004907159682643816214685929638952175999932299156089414639761565182862536979208272237582511852109168640000000000000000000000
100! = 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000
You should look for overflow, print the value after each iteration.
Note that rec(t); doesn't do anything as it doesn't use the returned value... you want t = rec(t);.
int is definitely too short, try long long... if that's still overflowing, you need another data structure.. eg: GMP Library.
Note: using some "proper" language for the job might give you some insight to the range you have to support... e.g. with python:
>>> import math
>>> math.factorial(100)
93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000L
private static void problem20()
{
string muliplent = "100";
for (int i = 99; i > 1; i--)
{
muliplent = getproduct(muliplent, i);
}
int sum = 0;
char[] result=muliplent.ToCharArray();
int count = muliplent.ToCharArray().Count();
for (int j = 0; j < count; j++)
{
sum = sum + (result[j] - '0');
}
Console.WriteLine("sum is {0}", sum);
Console.ReadLine();
}
private static string getproduct(string multiplent, int multiplier)
{
StringBuilder str = new StringBuilder();
int product = 0;
int remainder = 0;
int dividend = 0;
char[] c = multiplent.ToCharArray();
for (int i = c.Count() - 1; i >= 0; i--)
{
product = (((c[i] - '0') * multiplier) + dividend);
remainder = product % 10;
dividend = product / 10;
if (i != 0)
{
str.Insert(0, remainder);
}
}
str.Insert(0, product);
return str.ToString();
}