Sum of Consecutive Prime Numbers

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

题目大意呢，就是一个数如果有连续素数之和等于这个数，就加一次，最后输出这个数的次数.其实就是筛法就素数，再素数上在运算.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
int main()
{
    int i,j,n,cmp,w,prime[10005]= {0};
    prime[0]=prime[1]=1;
    for(i=2; i*i<=10000; i++)
        if(prime[i]==0)
            for(j=i*i; j<=10000; j+=i)
                prime[j]=1;  //筛法求素数打表
    while(scanf("%d",&n)!=EOF&&n)
    {
        cmp=w=0;
        for(i=2; i<=10000; i++)
            if(prime[i]==0)
            {
                cmp+=i;
                if(cmp==n)
                {
                    w++;
                    continue;
                }//别把这个数的本身忘掉....等于这个数了，在加素数肯定超出，故继续循环。
                for(j=i+1; j<=10000; j++)
                {
                    if(prime[j]==0)
                        cmp+=j;
                    if(cmp==n)
                    {
                        w++;
                        break;
                    }
                    if(cmp>n)
                    {
                        cmp=0;
                        break;
                    }
                }
            }
        printf("%d\n",w);
    }
    return 0;
}