I want to add a second loop to my program, in which I let the user enter an number x in each loop and form the sum over the entered numbers. the result of this code is 378 but must be for example 378+x=380
int main()
{
int i;
int sum = 0;
for (i = 1; i <= 27; i++) {
sum += i;
}
printf("%d\n", sum);
return 0;
}
You write a loop inside your loop and add a scanf.
int i, j;
int sum = 0;
for (int i = 0; i <= 27; i++) {
for (j = 0; j < 10; j++) {
int x= 0;
scanf ("%d", &x);
sum += x;
}
}
If you add a second loop you will need to input y*x inputs where y is the frequency of 2nd loop.
The code you may be looking for is
int main () {
int i,x;
int sum = 0;
for(i = 1; i <= 27; i++)
{
scanf("%d",&x);
sum += i+x;
}
printf("%d\n", sum);
return 0;
}
Related
*i want to copy inputed array in sum_of_elements function as argument and then sum all the elements of array, but i am getting output 0.
#include <stdio.h>
int i, num, sum;
int sum_of_elements(int arr[]) {
for (i = 0; i < num; i++) {
for (i = 0; sum = 0, i < num; i++) {
sum += arr[i];
}
return sum;
}
}
int main() {
printf("enter number of digits you want to add\n");
scanf("%d", & num);
int arr[num];
for (i = 0; i < num; i++) {
printf("enter number %d\n", i + 1);
scanf("%d", & arr[i]);
}
int total = sum_of_elements(arr);
printf("%d", total);
return 0;
Look at this line of code:
for (i = 0; sum = 0, i < num; i++) {
This resets sum to 0 every loop.
It should be
for (i = 0, sum = 0; i < num; i++) {
But it's probably better to do this:
sum = 0;
for (i = 0; i < num; i++) {
The issue was with the double for loop in your sum_of_elements function.
Removing the extra for loop, resolves the error.
#include <stdio.h>
int i, num, sum;
int sum_of_elements(int arr[]) {
for (i = 0; i < num; i++) {
sum += arr[i];
}
return sum;
}
int main() {
printf("enter number of digits you want to add\n");
scanf("%d", & num);
int arr[num];
for (i = 0; i < num; i++) {
printf("enter number %d\n", i + 1);
scanf("%d", &arr[i]);
}
int total = sum_of_elements(arr);
printf("%d", total);
return 0;
}
I have got some more problems with the code. This program ask the user to specify the nr of throws then it throws 3 dices and add these 3 dices to sum.
Then another function sorts the sum form the smallest to the largest with a bubble sorting algorithm.
the first two functions seems to work but the program does not print out the result of the 3rd sorting function.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define MAX 100
//This function ask the user for the amout of throws
int numberofthrows() {
int throws
printf("Type in the number of throws");
scanf("%d", &throws);
return throws;
}
//This function makes the random throws of 3 dices with regard to the number of throws
int filler(int thrownr, int dice1[MAX], int dice2[MAX], int dice3[MAX], int sum[MAX]) {
int i, nr;
srand(time(NULL));
for(i = 0; i <= thrownr; i++) {
nr = rand()%6;
dice1[i] = nr + 1;
nr = rand()%6;
dice2[i] = nr + 1;
nr = rand()%6;
dice3[i] = nr + 1;
sum[i] = dice1[i] + dice2[i] + dice3[i];
}
int j;
for(j = 0; j <= thrownr; j++) {
printf("%d ", dice1[j]);
printf("%d ", dice2[j]);
printf("%d ", dice3[j]);
printf("%d\n", sum[j]);
}
}
//This function sorts the result in form the sum array
int sorter(int thrownr, int sum[MAX], int sortsum[MAX]) {
int tmp, i, j, k, m;
for(i = 0; i <= thrownr; i++) {
sortsum[i] = sum[i];
}
for(m = 0; m <= 10; m++) {
for(j = 0; j <= thrownr; i++) {
if (sortsum[j] > sortsum[j+1]) {
tmp = sortsum[j];
sortsum[j] = sortsum[j+1];
sortsum[j+1] = tmp;
}
}
}
for(k = 0; k <= thrownr; k++) {
printf("%d\n", sortsum[k]);
printf("%d\n", sum[k]);
}
}
int main(void) {
srand(time(NULL));
int dice1[MAX];
int dice2[MAX];
int dice3[MAX];
int sum[MAX];
int sortsum[MAX];
int numberofthrows2;
numberofthrows2 = numberofthrows();
filler(numberofthrows2, dice1, dice2, dice3, sum);
sorter(numberofthrows2, sum, sortsum);
return 0;
}
The code for sorting is a bit wrong. Change
for(m = 0; m <= 10; m++)
To
for(m = 0; m <= thrownr-1; m++)
And
for(j = 0; j <= thrownr; i++)
To
for(j = 0; j < thrownr-m-1; i++)
To fix it. Also, call srand once at the start of main. Don't call it more than once in a program or you might get the same "random" numbers everytime you run your program.
QUESTION:
http://www.hpcodewars.org/past/cw17/problems/Prob02--CheckDigit.pdf
Here's my code:
int checkdigit(){
int n,i,j,sum1,sum2,k;
char ch;
printf("Enter the number of lines.Then enter ther the codes!");
scanf("%d",&n);
char *codes[n];
int msum[n];
int fsum[n];
for(i=0;i<n;++i){
scanf("%s",codes[i]);
}
for(i=0;i<n;++i){
for(j=0,k=3;j<21;j+=3,k+=3){
char *num;
num=codes[i];
ch=num[j];
sum1+=atoi(ch);
if(k<21)
ch=num[k];
sum2+=atoi(ch);
}
msum[i]=sum1*3;
fsum[i]=((msum[i]+sum2)%10);
if(fsum[i]!=0)
fsum[i]-=10;
}
for(k=0;k<sizeof(fsum);k++){
printf("%s %d",codes[k],fsum[k]);
}
return 0;
}
The Code now crashes after taking the first UPC code as input.
Change
fsum[i]=((msum+sum2)%10);
to
fsum[i]=((msum[i]+sum2)%10);
That is because msum is an array of integers, and msum[i] is an integer. As msum is an array, it is of type int* and is not compatible with int for the binary operator %
Here
char *codes[n];
is array of n pointers and you don't allocate memory for these pointers and try to scan values to this location so you see a crash
#include <stdio.h>
int checkdigit(int data[12]){
int i, even, odd, result;
even = odd = 0;
for(i = 0; i < 11; ++i){
if(i & 1)
even += data[i];
else
odd += data[i];
}
result = (odd * 3 + even) % 10;
if(result)
result = 10 - result;
return data[11] = result;
}
int main(){
int n;
scanf("%d", &n);
int codes[n][12];
int i, j;
for(i = 0; i < n; ++i){
for(j = 0; j < 11; ++j){
scanf("%d", &codes[i][j]);
}
checkdigit(codes[i]);
}
for(i = 0; i < n; ++i){
for(j = 0; j < 12; ++j){
if(j) putchar(' ');
printf("%d", codes[i][j]);
}
putchar('\n');
}
return 0;
}
I want to print this pattern like right angled triangle
0
909
89098
7890987
678909876
56789098765
4567890987654
345678909876543
23456789098765432
1234567890987654321
I wrote the following code:
#include <stdio.h>
#include <conio.h>
void main()
{
clrscr();
int i,j,x,z,k,f=1;
for ( i=10;i>=1;i--,f++)
{
for(j=1;j<=f;j++,k--)
{
k=i;
if(k!=10)
{
printf("%d",k);
}
if(k==10)
{
printf("0");
}
}
for(x=1;x<f;x++,z--)
{
z=9;
printf("%d",z);
}
printf("%d/n");
}
getch();
}
What is wrong with this code? When I check manually it seems correct but when compiled gives different pattern
Fairly simple: use two loops, one for counting up and one for counting down. Print literal "0" between the two.
#include <stdio.h>
int main()
{
for (int i = 0; i < 10; i++) {
for (int j = 10 - i; j < 10; j++)
printf("%d", j);
printf("0");
for (int j = 9; j >= 10 - i; j--)
printf("%d", j);
printf("\n");
}
return 0;
}
Like H2CO3's, but since we're only printing single digits why not use putchar():
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int i, j;
for(i = 0; i < 10; ++i)
{
// Left half.
for(j = 0; j < i; ++j)
putchar('9' - i + j + 1);
// Center zero.
putchar('0');
// Right half.
for(j = 0; j < i; ++j)
putchar('9' - i + j + 1);
putchar('\n');
}
return EXIT_SUCCESS;
}
Modified Code:
Check your errors:
# include<stdio.h>
# include<conio.h>
int main()
{
// clrscr();
int i,j,x,z,k,f=1;
for ( i=10;i>=1;i--,f++)
{
k=i; // K=i should be outside of loop.
for(j=1;j<=f;j++,k++)
{
if(k!=10)
{
printf("%d",k);
}
if(k==10)
{
printf("0");
}
}
z=9; //z=9 should be outside loop.
for(x=1;x<f;x++,z--)
{
printf("%d",z);
}
printf("\n");
}
//getch();
return 0;
}
You are defining k=i inside the for loop(loop which has j) so every time k gets value of i and thus it always get value of i and prints that value and your another condition(if(k==10)) will never be true because every time k takes value of i and i is less than 10 after first iteration of loop and z=9 inside loop so every time loop is executed it is taking value z=9 so it is printing wrong value.
Here's a C# version:
static void DrawNumberTriangle()
{
for (int line = 10; line >=1; line--)
{
for (int number = line; number < 10; number++)
{
System.Console.Write(number);
}
System.Console.Write("0");
for (int number = 9; number > line - 1; number--)
{
System.Console.Write(number);
}
System.Console.WriteLine();
}
}
I'd suggest renaming your i,j,x,z,k,f variables to ones that have meaning like the one's I used. This helps making your code easier to follow.
Rather than output the mid 0 using printf, why not print it using the loops itself.
The following short and simple code can be used:
int main()
{
int m = 10, n, p;
while(m >= 1)
{
for(n = m; n <= 10; n++)
printf("%d", n % 10);
for(p = n - 2; p >= m; p--)
printf("%d", p );
printf("\n");
m--;
}
return 1;
}
For high throughput (though of questionable merit in terms of clarity):
#include <stdio.h>
int main() {
char const digits[] = "1234567890";
char const rdigits[] = "9876543210";
for (int i = 0; i < 30; ++i) {
int k = i % 10;
fputs(digits + 9 - k, stdout);
for (int j = 9; j < i; j += 10) fputs(digits, stdout);
for (int j = 9; j < i; j += 10) fputs(rdigits, stdout);
fwrite(rdigits, 1, k, stdout);
fputs("\n", stdout);
}
}
#include <stdio.h>
void print(int i){
if(i == 10){
putchar('0');
return ;
} else {
printf("%d", i);
print(i+1);
printf("%d", i);
}
}
int main(void){
int i;
for(i = 10; i>0; --i){
print(i);
putchar('\n');
}
return 0;
}
I want to print this pattern like right angled triangle
0
909
89098
7890987
678909876
56789098765
4567890987654
345678909876543
23456789098765432
1234567890987654321
I wrote the following code:
#include <stdio.h>
#include <conio.h>
void main()
{
clrscr();
int i,j,x,z,k,f=1;
for ( i=10;i>=1;i--,f++)
{
for(j=1;j<=f;j++,k--)
{
k=i;
if(k!=10)
{
printf("%d",k);
}
if(k==10)
{
printf("0");
}
}
for(x=1;x<f;x++,z--)
{
z=9;
printf("%d",z);
}
printf("%d/n");
}
getch();
}
What is wrong with this code? When I check manually it seems correct but when compiled gives different pattern
Fairly simple: use two loops, one for counting up and one for counting down. Print literal "0" between the two.
#include <stdio.h>
int main()
{
for (int i = 0; i < 10; i++) {
for (int j = 10 - i; j < 10; j++)
printf("%d", j);
printf("0");
for (int j = 9; j >= 10 - i; j--)
printf("%d", j);
printf("\n");
}
return 0;
}
Like H2CO3's, but since we're only printing single digits why not use putchar():
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int i, j;
for(i = 0; i < 10; ++i)
{
// Left half.
for(j = 0; j < i; ++j)
putchar('9' - i + j + 1);
// Center zero.
putchar('0');
// Right half.
for(j = 0; j < i; ++j)
putchar('9' - i + j + 1);
putchar('\n');
}
return EXIT_SUCCESS;
}
Modified Code:
Check your errors:
# include<stdio.h>
# include<conio.h>
int main()
{
// clrscr();
int i,j,x,z,k,f=1;
for ( i=10;i>=1;i--,f++)
{
k=i; // K=i should be outside of loop.
for(j=1;j<=f;j++,k++)
{
if(k!=10)
{
printf("%d",k);
}
if(k==10)
{
printf("0");
}
}
z=9; //z=9 should be outside loop.
for(x=1;x<f;x++,z--)
{
printf("%d",z);
}
printf("\n");
}
//getch();
return 0;
}
You are defining k=i inside the for loop(loop which has j) so every time k gets value of i and thus it always get value of i and prints that value and your another condition(if(k==10)) will never be true because every time k takes value of i and i is less than 10 after first iteration of loop and z=9 inside loop so every time loop is executed it is taking value z=9 so it is printing wrong value.
Here's a C# version:
static void DrawNumberTriangle()
{
for (int line = 10; line >=1; line--)
{
for (int number = line; number < 10; number++)
{
System.Console.Write(number);
}
System.Console.Write("0");
for (int number = 9; number > line - 1; number--)
{
System.Console.Write(number);
}
System.Console.WriteLine();
}
}
I'd suggest renaming your i,j,x,z,k,f variables to ones that have meaning like the one's I used. This helps making your code easier to follow.
Rather than output the mid 0 using printf, why not print it using the loops itself.
The following short and simple code can be used:
int main()
{
int m = 10, n, p;
while(m >= 1)
{
for(n = m; n <= 10; n++)
printf("%d", n % 10);
for(p = n - 2; p >= m; p--)
printf("%d", p );
printf("\n");
m--;
}
return 1;
}
For high throughput (though of questionable merit in terms of clarity):
#include <stdio.h>
int main() {
char const digits[] = "1234567890";
char const rdigits[] = "9876543210";
for (int i = 0; i < 30; ++i) {
int k = i % 10;
fputs(digits + 9 - k, stdout);
for (int j = 9; j < i; j += 10) fputs(digits, stdout);
for (int j = 9; j < i; j += 10) fputs(rdigits, stdout);
fwrite(rdigits, 1, k, stdout);
fputs("\n", stdout);
}
}
#include <stdio.h>
void print(int i){
if(i == 10){
putchar('0');
return ;
} else {
printf("%d", i);
print(i+1);
printf("%d", i);
}
}
int main(void){
int i;
for(i = 10; i>0; --i){
print(i);
putchar('\n');
}
return 0;
}