数学趣题——验证尼克彻斯定理

任何一个整数的立方都可以表示成一串连续的奇数的和。

   1: #include "stdio.h"

   2:  

   3: void Nicoqish(int N)

   4: {

   5:     int i, j, sum = 0;

   6:  

   7:     for(i = 1; i < N * N * N; i = i + 2)         /*i为起点*/

   8:         for(j = i; j < N * N * N; j = j + 2)         /*j控制从i向后顺次累加*/

   9:         {

  10:             sum = sum + j;

  11:  

  12:             if(sum == N * N * N)

  13:             {

  14:                 printf("%d=%d+%d...+%d\n", N * N * N, i, i + 2, j);

  15:                 return;

  16:             }

  17:  

  18:             if(sum > N * N * N)

  19:             {

  20:                 sum = 0;

  21:                 break;

  22:             }

  23:         }

  24: }

  25:  

  26: int main()

  27: {

  28:     int N;

  29:     printf("Please input a integer to verify Nicoqish Law\n");

  30:     scanf("%d", &N);

  31:     Nicoqish(N);

  32:     return 0;

  33: }