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Read integers a and b (b ≥ 0) and print a raised to the power b.
Anything to the power 0 is 1. A loop that multiplies is enough — you do not need pow from math.h (that returns a double).
Input. Two integers a and b.
Output. One integer: a^b.
Constraints
- 0 ≤ b ≤ 12
- Results fit in a 32-bit int.
Examples
Input
2 10
Output
1024
Input
5 0
Output
1
Hint
- long result = 1; for (int i = 0; i < b; i++) result *= a;
Show correct code
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#include <stdio.h>
int main(void) {
int a, b;
if (scanf("%d %d", &a, &b) != 2) return 1;
long result = 1;
for (int i = 0; i < b; i++) result *= a;
printf("%ld\n", result);
return 0;
}
main.cC17 · gcc · Ctrl + Enter runs
ResultIdle
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