I am looking for a narcissistic number with 3 to 9 digits. I have working code, but the use of nested loops is cumbersome. I think I could use recursion, how?
#include "stdafx.h"
#include <time.h>
int power(int a,int b)
{
int t=1,i;
for (i=0;i<b;i++)
t*=a;
return t;
}
int _tmain(int argc, _TCHAR* argv[])
{
double t0=clock();
int a,b,c,d,e,f,g,h,i;
int len=9;
for (a=1;a<=9;a++)
for (b=0;b<=9;b++)
for (c=0;c<=9;c++)
for (d=0;d<=9;d++)
for (e=0;e<=9;e++)
for (f=0;f<=9;f++)
for (g=0;g<=9;g++)
for (h=0;h<=9;h++)
for(i=0;i<=9;i++)
{
int num=10*(10*(1000000*a+100000*b+10000*c+1000*d+100*e+10*f+g)+h)+i;
if (num==power(a,len)+power(b,len)+power(c,len)+power(d,len)+power(e,len)+power(f,len)+power(g,len)+power(h,len)+power(i,len))
printf(" %d ",num);
}
printf("\n耗时： %f s",(clock()-t0)/CLOCKS_PER_SEC);
return 0;
}
Your num is simply iterating through all numbers from 0 (See Edit 2) to 999,999,999. So you really don't need to iterator over every digit. A simple way of doing what you want, even though probably not anymore efficient is this:
unsigned int num;
for (num = 0; num < 1000000000; ++num) // will take long
{
char digits[15];
unsigned int i, other_num;
sprintf(digits, "%09u", i); // division used here
other_num = 0;
for (j = 0; j < 9; ++j)
other_num += power(digits[j] - '0', len);
if (num == other_num)
printf(" %d ",num);
}
Note that you can do your power function in a more efficient way. See this question.
Edit 1: (Note: see Edit 2)
I am reading through the wikipedia page, and it turns out that you shouldn't set len to 9, because if you have for example 153, the other_num should be 1^3 + 5^3 + 3^3 (example from Wikipedia).
So you should adjust the loop in the following way:
unsigned int num;
for (num = 0; num < 1000000000; ++num)
{
char digits[15];
unsigned int i, other_num, len;
len = snprintf(digits, 10, "%u", i); // len returned by snprintf
other_num = 0;
for (j = 0; j < len; ++j) // changed 9 to len
other_num += power(digits[j] - '0', len);
if (num == other_num)
printf(" %d ",num);
}
Edit 2:
I just noticed that your number starts from 100,000,000, so you don't really need to pay attention to Edit 1. Nevertheless, I believe it's good to know it in case you needed to change the range of your numbers.
Wonderful loop.
What about this:
for( num = 100000000; num <= 999999999; num++ ) {
int compare = 0;
for( k = 0; k < 10; k++ ) {
int position = pow(10, k+1);
int s = num%position;
compare += pow(s, 9);
}
if( num == compare ) {
printf("%d\n", num);
}
}
Of course, check for boundaries :)
Actually, you are inspecting all values (num) from 1 to 10^len, if I am not mistaken.
So, a simple for cycle from 1 to 10^len would be enough. Then you can take the digits from num and calculate the power value.
I would strongly suggest to calculate the powers 1^len, 2^len ... 9^len first and put them into an array. It will greatly improve performance.
Related
I've been trying to do this problem on Project Euler
This is what I've done so far -
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main(){
char nums[] = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450"
int64_t sum = 1, new_sum = 0;
int num;
for(int i = 0; i < 988; i+= 13){
sum = 1
for(int j = 0; j < 13; ++j){
//Converting nums[i] to int and storing it to num here
sum *= num;
}
if(sum > new_sum){
new_sum = sum;
}
}
printf("%lld", new_sum);
}
I don't know how to convert each charNum in the String to an integer, I've tried atoi and sscanf and both of them ask for a char pointer.
I get these errors respectively
passing argument to parameter here
int atoi(const char *); //sum *= atoi(nums[i])
format specifies type 'int *' but the argument has type 'int' [-Wformat]
sscanf(nums[i], "%d", num);
Any help is appreciated.
Just converting a character representing a digit to the digit itself is simply done by
int digit = nums[position] - '0';
C language guarantees for any character set that the digits 0-9 are succeeding one another in exactly that order (which is not necessarily the case for alphabetic characters, see e.g. (in-?) famous EBCDIC).
As I read the problem the maximum can be at any location, not only multiples of 13. For this speaks as well that the maximum of four subsequent digits is found at a location where the number of digits preceding is not a multiple of four.
Apart from, unless the number if digits is a multiple of 13, your loops would exceed array bounds of nums, so undefined behaviour (if not a multiple of 13, then for last iteration of i there are less than 13 digits left, but j still tries to iterate over all 13 digits).
Fixing both:
size_t sequenceLength = 13; // could be a function parameter
size_t len = strlen(nums); // sizeof would include the terminating 0 character!
if(len < sequenceLength)
{
return 0; // if within a function, else whatever appropriate error handling
}
uint64_t max = 1; // can't have negative values anyway...
for(size_t i = 0; i < sequenceLength; ++i)
{
max *= nums[i] - '0'; // see above
}
// now we found first possible maximum: the product of first 13 digits
uint64_t num = max;
for(size_t i = sequenceLength; i < len; ++i)
{
// for next product, we have already included the first 12 digits
// within the previous product!
// but there is one surplus digit contained we need to eliminate:
num /= nums[i - sequenceLength] - '0';
// now we can include the yet missing one:
num *= nums[i] - '0';
// or as a one-liner:
num = num / (nums[i - sequenceLength] - '0') * (nums[i] - '0');
// TODO: update maximum, if need be, just as in your code
}
Would you please try the following:
#include <stdio.h>
#include <stdlib.h>
#define N 13
int main() {
char nums[] = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";
int64_t prod = 1, max_prod = 0;
int num;
int pos;
for (int i = 0; i < sizeof nums; i += N) {
prod = 1;
for (int j = 0; j < N; j++){
// Converting nums[i] to int and storing it to num here
num = nums[i + j] - '0';
prod *= num;
}
if (prod > max_prod) {
max_prod = prod;
pos = i;
}
}
printf("max product = %ld at %.*s\n", max_prod, N, nums + pos);
}
Output:
max product = 6270566400 at 4355766896648
BTW as the variable name sum is misleading, I've changed it to prod for product.
I assume that you want a nuber for each digit (i.e., range 0 to 9)
If yes, the below code should be not too far from what you need:
char nums[] = "73167176531330624919225119674426574742355349194934969835203... (so long)...";
char *p;
long lTheTotalCount, lTheNumber;
lTheTotalCount = 0;
for (p=nums ; *p ; p++) {
lTheNumber = (long)(*p - '0');
lTheTotalCount += lTheNumber;
};
Sample Input
2
2
5
Sample Output
0 1 1
0 1 1 2 1 2
I already knew how to transfer an integer to its binary and count how many 1 in its binary.
But my code only can input one integer each time. I want it to input many numbers, like the sample input and sample output. To make question more easy to understand, so I drew a picture. Thank you!!
Thanks all of you!! But I need some time to understand those code and my English is pretty basic, so I couldn't reply you guys soon. But I will understand and reply you as soon as possible!!thank you :D
#include <stdio.h>
int main()
{
int n,cnt=0,m;
scanf("%d",&n);
while(n>0){
m=n%2;
if(m==1){
cnt++;
}
n/=2;
}
printf("%d",cnt);
return 0;
}
In practice, the process can be simplified and accelerated by noticing that if we know the result for i < 2^p, then, for all values in [2^p, 2^(p+1)-1], we have
count(j) = 1 + count(j-2^p)
This method is useful here as we have to provide the result for all values less or equal to n.
Moreover, in order to avoid performing the same calculation at different times, we first calculate the maximum n value, and perform the calculation for this maximum value.
#include <stdio.h>
int main() {
int t;
int check = scanf("%d", &t);
if (check != 1) return 1;
int nn[t];
for (int i = 0; i < t; ++i) {
int check = scanf("%d", &nn[i]);
if (check != 1) return 1;
}
int nmax = 0;
for (int i = 0; i < t; ++i) {
if (nn[i] > nmax) nmax = nn[i];
}
int count[nmax+1];
count[0] = 0;
count[1] = 1;
int pow2 = 1;
do {
pow2 *= 2;
for (int j = pow2; (j < 2*pow2) && (j <= nmax); j++) {
count[j] = 1 + count[j-pow2];
}
} while (pow2 <= nmax+1);
for (int i = 0; i < t; ++i) {
for (int j = 0; j <= nn[i]; ++j) {
printf ("%d ", count[j]);
}
printf ("\n");
}
return 0;
}
Let the number of measured values be t. Now if you want to take t number of integers as input, you have to run a loop t times and perform necessary tasks (like taking value of n, calculating cnt etc) inside that loop. Check the following code snippet:
int main()
{
int n,cnt=0,m;
int numberOfMeasuredValues;
scanf("%d", &numberOfMeasuredValues);
while(numberOfMeasuredValues > 0){
cnt = 0;
scanf("%d",&n);
// perform necessary calculations
printf("%d\n",cnt);
numberOfMeasuredValues--;
}
return 0;
}
In the code numberOfMeasuredValues is the t I've used in the explanation.
You can pre-allocate number of cases and work on them with outer and inner loops. I've added the code with comments below.
#include <stdio.h>
int main(void)
{
int cases; // Number of cases
scanf("%d", &cases);
int* nums = malloc(cases * sizeof *nums); // Allocate enough space.
/* Take the inputs. */
for (int i = 0; i < cases; i++)
{
scanf("%d", (nums + i));
}
int cnt = 0; // Counter for number of 1s.
// Outer loop for going through cases.
for (int i = 0; i < cases; i++)
{
// Inner loop for couting towards inputted number.
for (int j = 0; j < *(nums + i) + 1; j++)
{
/*
* You can Implement this part by yourself if you wish.
*/
for (int k = 0; k < 32; k++)
{
if ((j >> (31 - k)) & 0x01) cnt++; // Increment counter if digit is 1.
}
printf("%d ", cnt);
cnt = 0;
}
printf("\n");
}
return 0;
}
Here's the following code to your problem
#include<iostream>
using namespace std;
int main()
{
int tcs;
cin >> tcs;
while (tcs--) {
int n;
cin >> n;
for (int i = 0 ; i <= n ; i++) {
cout << __builtin_popcount(i) << " ";
}
cout << "\n";
}
return 0;
}
What I want to do is to get a cumulative sum of previous integers starting from 1, for example:
If my input is 4, then the function should work in this way;
1 + (1+2) + (1+2+3) + (1+2+3+4) = 20.
And the output needs to be 20. Also, I have to get this done by a function, not in main(); function while using int n as the only variable.
What I've tried is to make a function which adds from 1 to integer N, and use 'for'to make N start from 1, so that it can fully add the whole numbers until it reaches N.
#include <stdio.h>
int sum(int n);
int main() {
int n, input, sum;
sum = 0;
scanf("%d", &n);
for (n = 0; n <= input; n++) {
sum += n;
}
printf("%d", sum);
return 0;
}
int sum(int n) {
int i, n, sum = 0;
scanf("%d", &n);
for (i = 1; i <= n; i += 1){
sum += i;
}
return n;
}
What I expected when the input is 4 is 20, but the actual output is 10.
I would have written it this way, remarks are where changes been made
#include <stdio.h>
int sum(int n);
int main() {
int n, input, sum;
// sum = 0; // no need for this
scanf("%d", &n);
/* the next for has no use
for (n = 0; n <= input; n++) {
sum += n;
} */
// I would be adding some input sanitazing if possible here
printf("%d", sum(n));
return 0;
}
int sum(int n) {
int i, /*n, */ rsum = 0; // n is already a parameter, rsum for running sum
// scanf("%d", &n); // nope nope, scanf and printf should be avoided in functions
for (i = 1; i <= n; i++){ // changed i +=1 with i++ , easier to read
for (j=1;j<=i;j++) // need this other loop inside
rsum += j;
}
return rsum;
}
Here it is with a single loop; very fast.
#include <stdio.h>
int cumulative_sum(int m)
{
int sum = 0;
for(int n=1; n<=m; ++n) sum += n*(n+1);
return sum/2;
}
int main(void)
{
int n;
printf("Input value N: ");
scanf("%d", &n);
printf("Answer is %d\n", cumulative_sum(n));
return 0;
}
The main issue is in the function, you are doing only 1 loop (you have also some logical things, which compiler should tell you, like same naming of variable and function.. eg.),
so in case you will put 4 as the input, loop will do only 1+2+3+4, but your case if different, you want to make suma of all iterations like 1 + (1+2) + (1+2+3) + (1+2+3+4)
you are doing only last step basically (1+2+3+4), 4 iterations (4x suma), but actually you need 10 iterations (due suma of all particular suma of elements)
As suggested, try to debug your code - What is a debugger and how can it help me diagnose problems?
- it will really help you do understand your code
As mentioned, the issue is in
int sum(int n) {
int i, n, sum = 0;
scanf("%d", &n);
for (i = 1; i <= n; i += 1){
sum += i;
}
return n;
}
You have to make two loops eg. like follows:
int sum,n = 0;
//scanf("%d", &n);
n = 4; //input simulation
//just for demonstration
int iterations = 0;
//counter for total loops (how many "brackets" needs to be count)
for(int loopsCounter = 1; loopsCounter <= n;loopsCounter++){
//counter for child elements in brackets (like 1+2 ,1+2+3, ..)
for (int sumaLoopCounter = 1; sumaLoopCounter <= loopsCounter; sumaLoopCounter++){
//simply make sum with the given number
/* first step 0 +1
second 1+2 added to previous suma = 1+3
third 1+2+3,.. added to previous = 4+6
...
*/
sum += sumaLoopCounter;
//just testing for demonstration
iterations++; //iterations = iterations + 1
}
}
printf("%i \n",iterations);
printf("%i",sum);
Then you got output as expected - sum of all "bracket elements" and 10 iterations, which matches numbers of needed additions
10
20
I tackled the problem by first figuring out the length of two given numbers and aligning the one with less digits (if one exists) into a new array so that the ones, tens, hundreds etc. align with the bigger number's ones, tens, hundreds, etc.
Then I wanted to save the sum of each two aligned elements (with a mod of 10) into a new array while checking if the sum of digits is greater than 10 - just the basic sum stuff. Now the problem occurs with adding two elements into the aplusb integer and I've tried fixing it with writing
int aplusb = (lengthA[max-i]-'0') +(temp[max-i]-'0');
but it doesn't work. I'm stuck and I don't know what to do. Please help.
The whole code:
#include <stdio.h>
#include <math.h>
int main(){
char a[10000];
char b[10000];
scanf("%s %s", &a, &b);
char sum[10000];
int lengthA = 0;
int lengthB = 0;
int i = 0;
while(a[i]){
i++;
} lengthA = i;
i = 0;
while(b[i]){
i++;
} lengthB = i;
char temp[10000];
int aplusb;
int carry = 0;
int max = lengthA;
int difference = abs(lengthA - lengthB);
if(lengthA>lengthB){
for(i=0; i<lengthA; i++){
temp[i+difference]=b[i];
}
for(i=0; i<=max; i++){
aplusb = lengthA[max-i]+temp[max-i]; //<-- this is the problematic line
if(carry = 1) aplusb++;
if(aplusb>9){
carry = 1;
aplusb%=10;
}
sum[i]=aplusb;
}
}
for(i=0; i<=max; i++){
printf("%c", sum[i]);
}
/*
if(lengthB>lengthA){
max = lengthB;
for(i=0; i<lengthB; i++){
temp[i+difference]=a[i];
}
}*/
return 0;
}
Doing operations and storing on very large numbers is very akin to doing operations and storing polynomials, i.e. with x = 10. a0 + a1.10 + a2.10^2 ... + an.10^n.
There are many polynomial libraries on the Internet, where you could find inspiration. All operations on your very large numbers can be expressed in terms of polynomials. This means that by using base 2^8, or even base 2^63, instead of base 10 to internally store your large numbers you would greatly improve performance.
You must also normalize your coefficients after operations to keep them positive. Operations may result in a negative coefficient, That can easily be fixed, as it is very similar to borrowing after a subtraction, this means coefficients must be larger than your base by 1bit.
To convert back to base 10, you'd need to solve r (your result) for v (your value), such as r(10)=v(2^63). This has only one solution, if you enforce the positive coefficients rule.
[note] After thinking about it some more: the rule on positive coefficients may only be necessary for printing, after all.
Example: adding. no memory error checking
int addPolys(signed char** result, int na, const signed char* a, int nb, const signed char* b)
{
int i, nr, nmin, carry, *r;
nr = max(na, nb) + 1;
nmin = min(na, nb);
r = malloc(sizeof(signed char) * (na + nb + 1));
if (nb < na)
{
nr = nb;
}
for (i = 0; i < nmin; ++i)
{
r[i] = a[i] + b[i];
}
for (; i < na; ++i)
{
r[i] = a[i];
}
for (; i < nb; ++i)
{
r[i] = b[i];
}
r[nr - 1] = 0;
// carry - should really be a proc of its own, unoptimized
carry = 0;
for (i = 0; i < nr; ++i)
{
r[i] += carry;
if (r[i] > 10)
{
carry = r[i] / 10;
r[i] %= 10;
}
else if (r[i] < 0)
{
carry = (r[i] / 10) - 1;
r[i] -= (carry * 10);
}
else
carry = 0;
}
// 'remove' leading zeroes
for (i = nr - 1; i > 0; --i)
{
if (r[i] != 0) break;
}
++i;
*result = r;
if (i != nr)
{
*result = realloc(i * sizeof(signed char));
}
return i; // return number of digits (0 being 1 digit long)
}
That code is working now for any two positive numbers with up to ten thousand digits:
#include <stdio.h>
#include <math.h>
#include <string.h>
int main(){
char chara[10000];
char charb[10000];
scanf("%s %s", &chara, &charb);
int lengthA = strlen(chara);
int lengthB = strlen(charb);
int max = lengthA;
if(lengthB>lengthA) max=lengthB;
int dif = abs(lengthA - lengthB);
//ustvari int tabele
int a[max];
int b[max];
int sum[max+1];
// nastavi nule
int i;
for(i=0; i<max; i++){
a[i] = 0;
b[i] = 0;
sum[i] = 0;
} sum[max] = 0;
//prekopiraj stevila iz char v int tabele &obrni vrstni red
for(i=0; i<lengthA; i++){
a[i] = chara[lengthA-i-1]-'0';
}
for(i=0; i<lengthB; i++){
b[i] = charb[lengthB-i-1]-'0';
}
int vsota;
int prenos = 0;
for(i=0; i<max; i++){
vsota = a[i]+b[i] + prenos;
if(vsota>=10) prenos = 1;
else if (vsota<10) prenos = 0;
sum[i]=vsota%10;
}
if(prenos==1){
sum[max] = 1;
for(i = max; i>=0; i--){
printf("%d", sum[i]);
}
} else {
for(i = max-1; i>=0; i--){
printf("%d", sum[i]);
}
}
return 0;
}
I have this assignment for my intro to C programming class and part of my code has to find the sequence of the sum of square digits of a number in order to determine after if the given number is a happy number (sum of square digits = 1)
Here's part of my code:
#include <stdio.h>
#include <math.h>
// The sum of square digits function
int sqd (int x) {
int sum = 0;
while (x > 0) {
sum = sum + pow(x%10, 2);
x = x/10;
}
return sum;
}
// The search function
int search (int a[], int val, int size) {
int i;
for (i = 0; i < size; i++) {
if (a[i] == val) {
return 1;
}
}
return 0;
}
// The main program
void main () {
int a [1000] = {0};
int N;
int count = 1;
int j;
printf("Please enter the potential happy number:\n", N);
scanf ("%d", &N);
a[0] = N;
a[count] = sqd (N);
do {
a[count] = sqd (a[count-1]);
count++;
} while (search (a, a[count], count));
for ( j = 0; j <= count; j++) {
printf("%d\n", a[j]);
}
}
It only prints the first three sums in the sequence. I really don't know how to make it work.
Thank you in advance
This line
while (search (a, a[count], count));
makes sure that you break out of the loop after one round since a[1] is not equal toa[0]. You can change that line to be:
while (a[count-1] != 1);
You also need to add a clause to make sure that you stop when the limit of the array is reached. Update that line to be:
while (a[count-1] != 1 && count < 1000 );
And then, change the printing loop to use i < count, not i <= count. Using <= will result in accessing the array out of bounds when the user enters a sad number.
for ( j = 0; j < count; j++){
printf("%d\n", a[j]);
}
Update
After a bit of reading on happy numbers at Wikipedia, I understand why you had call to search in the conditional of the while. The following also works.
} while ( ! (a[count-1] == 1 || search(a, a[count-1], count-1)) );
That will search for the last number in the array but only up to the previous index.