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  1. #include <stdio.h>
  2. int main(void)
  3. {
  4.     int m, n, i, j, sum, x;
  5.     printf("1^m + 2^m + 3^m + ... + n^m を求める\n");
  6.     printf("m=");scanf("%d", &m);
  7.     printf("n=");scanf("%d", &n);
  8.     sum = 0;
  9.     for (i = 1; i <= n; i++) {
  10.         for (x = j = 1; j <= m; j++) {
  11.             x *= i;
  12.         }
  13.         sum += x;
  14.     }
  15.     printf("%d\n", sum);
  16.     return 0;
  17. }
  18.  
Success #stdin #stdout 0s 2252KB
comments (?)
stdin 
3
4
stdout 
1^m + 2^m + 3^m + ... + n^m を求める
m=n=100
https://ideone.com/Elx3J3
language:
C (gcc 8.3)
created:
10 years ago
visibility:
public

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