Problem D: Prime Numbers…Again!

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本文探讨了一道算法题目，即如何计算一个正整数可以由多少种不同的连续素数之和的方式表示出来。通过预处理素数并使用滑动窗口的方法来解决这个问题。

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Problem D: Prime Numbers…Again!

Time Limit: 1 Sec Memory Limit: 128 MB

Description

We can write some of the numbers as the sum of some distinct consecutive prime numbers.

For instance, number 15 can be represented as 3+7+5=15

You are given a postive integer number 1<=n<=100000

and you should find the number of different ways that number n

can be represented as the sum of some distinct consecutive prime numbers.

Input

The number of test cases comes in the first line. For each test case you are given a positive integer

1<=n<=100000

Output

For each test case, print the number of the different ways that n

can be represented as a sum of some consecutive prime numbers.

Sample Input

Sample Output

#include<cstring>
#include<cstdio>
#include<iostream>
#define max 10000           //要注意这个数字，这个数字其实大小应该为100000的平方根，是根据一个数的因子小于等于这个数的平方根所得到的
using namespace std;

bool p[max]={0,0,1,1,0,1};
int prim[max]={2};
void deal()
{    int j,k=2;
     memset(p,true,sizeof(p));
     for(j=2;j<max;j++)
     {
         if(p[j])
         for(k=j*j;k<max;k+=j)
            p[k]=0;
     }
     k=0;
     for(j=2;j<max;j++)
        if(p[j])
        prim[k++]=j;
 }
 int main()
 {
     //freopen("1.in","r",stdin);
     //freopen("1.in","w",stdout);
     deal();
     int t,n,sum,i,j,way;
     scanf("%d",&t);
     while(t--)
     {
    	way=0;
    	scanf("%d",&n);
    	for(i=0;prim[i]<=n;i++)
    	{
    		sum=0;
	    	for(j=i;sum<=n;j++)
	    	{

	    		if(sum==n)
	    		{
		    		way++;
		    		break;
		    	}
		    	sum+=prim[j];
	    	}
	    }
	    printf("%d\n",way);
    }
     return 0;
 }