Sum of Consecutive Prime Numbers III

Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representations does a given positive integer have? For example, the integer 53 has two representations 5 + 7 + 11 + 13 + 17 and 53. The integer 41 has three representations 2+3+5+7+11+13, 11+13+17, and 41. The integer 3 has only one representation, which is 3. The integer 20 has no such representations. Note that summands must be consecutive prime
numbers, so neither 7 + 13 nor 3 + 5 + 5 + 7 is a valid representation for the integer 20.
Your mission is to write a program that reports the number of representations for the given positive integer.

输入

The input is a sequence of positive integers each in a separate line. The integers are between 2 and 10 000, inclusive. The end of the input is indicated by a zero.

输出

The output should be composed of lines each corresponding to an input line except the last zero.

严格按样例的格式输出。

与每个输入数据相对应，首先输出"Integer XXX"（XXX代表输入的整数），然后看是否有解，如果有解，按样例格式逐行输出各个解（多行解的输出顺序是，加式中开头数字越小的排得越前）；如果无解，输出"No Answer"

样例输入

5341340

样例输出

Integer 53----Answer 1: 5+7+11+13+17----Answer 2: 53Integer 41----Answer 1: 2+3+5+7+11+13----Answer 2: 11+13+17----Answer 3: 41Integer 3----Answer 1: 3Integer 4----No Answer

#include<stdio.h>#include<math.h>int main(){int i, j, k,z, a, b, c,d,n, s, sum[5010],e[5010], step;while (scanf("%d", &a) != EOF){if (a == 0) break;i = 0; j = 0;for (n = 2; n <= a; n++){for (s = 2; s*s <= n; s++)  //rt的算法if (n%s == 0)break;if (s*s>n){sum[i] = n;//   记录所有素数i++;j = i;}}b = 0; step = 0; k = 0; d = 0;while (j){i = k;for (; sum[i] != '\0'; i++){step = step + sum[i];if (step == a){e[d] = k;e[d + 1] = i;//  sum[k]...sum[i]b++;d+=2;k++;  step = 0;break;}if (step>a){k++;step = 0;break;}}j--;}if (b == 0){printf("Integer %d\n",a);printf("----No Answer\n");}if (b>0){printf("Integer %d\n", a);   //有答案n = 1 ;for (i = 0,z=0; z<b;z++, i+=2)  //每一次循环代表以此数据{j = e[i];  //  sum[j]k = e[i + 1]; //  sum[k]printf("----Answer %d: ", n);if (j<k){for (; j <=k; j++)//大于一个数的一组数据{printf("%d", sum[j]);if (j < k)  printf("+");}printf("\n");}if (j == k)//{printf("%d",a);printf("\n");}n++;}}//printf("%d\n", b);   组数}return 0;}