C program for power of an integer
```c
#include <stdio.h>
void main () {
int i;
int p;
int q;
int a;
printf("saisir l'entier");
scanf("%d", &a);
printf("saisir la puissance");
scanf("%d", &p);
if (p == 0)
q = 1;
else {
q = 1;
for (i = 1; i <= p; i++)
q = q * a;
}
printf("%d\n", q);
system("pause");
}
```</stdio.h>
#include <stdio.h>
void main () {
int i;
int p;
int q;
int a;
printf("saisir l'entier");
scanf("%d", &a);
printf("saisir la puissance");
scanf("%d", &p);
if (p == 0)
q = 1;
else {
q = 1;
for (i = 1; i <= p; i++)
q = q * a;
}
printf("%d\n", q);
system("pause");
}
```</stdio.h>
13 answers
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ContributorGood evening,
You can simplify the algorithm.
In fact, just need to set the variable q to 1 and use a simple while loop. There's no need to consider the case where the exponent is zero.
For example:#include <stdio.h> int main(void) { int a; int p; int q; printf("enter the integer: "); fflush(stdout); scanf("%d",&a); printf("enter the power: "); fflush(stdout); scanf("%d",&p); q=1; while(p--) q*=a; printf("%d\n",res); return 0; }
If you have any questions about the rest of the code, feel free to ask.
Best regards,
--
Google is your friend-
```c
#include <stdio.h>
#include <math.h>
int puiss(int x, int y)
{
int q;
int i = 0;
if (y == 0)
q = 1;
else
{
q = 1;
while (i < y)
{
q = q * x;
i++;
}
}
return q;
}
int main()
{
int a;
int p;
int waw;
printf("****Bonjour malek****");
printf("comment cv pas !");
printf("\n saisir l'entier x :\n ");
scanf("%d", &a);
printf("\n saisir la puissance p : \n ");
scanf("%d", &p);
/*calcule de puissance*/
waw = puiss(a, p);
printf("\n la puissance de %d ** %d est : %d \n", a, p, waw);
return 0;
}
```</math.h></stdio.h>
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