poj2739尺取法+素数筛

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.

Input

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.

Output

The output should be composed of lines each corresponding to an input line except the last zero. An output line includes the number of representations for the input integer as the sum of one or more consecutive prime numbers. No other characters should be inserted in the output.

Sample Input

2
3
17
41
20
666
12
53
0

Sample Output

1
1
2
3
0
0
1
2
题意：给你一个数，要求找出它能用连续素数相加而成的个数
题解：一看就知道要先来一个素数筛啦。然后用另一个数组保存2到10000的尺取结果，输入后就能直接输出了，刚开始还担心会不会TLE@-@，结果居然只花了188ms
果然还是打表大法好啊

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007

using namespace std;

const int N=20000+5,maxn=10000+5,inf=0x3f3f3f3f;

int p[maxn],ans[maxn];
bool isprime[N];

void getprime()
{
    int k=0;
    for(int i=0;i<N;i++)isprime[i]=1;
    isprime[0]=isprime[1]=0;
    for(int i=2;i<N;i++)
    {
        if(isprime[i])
        {
            p[k++]=i;
            for(int j=2*i;j<N;j+=i)
                isprime[j]=0;
        }
    }
}
int solve(int x)//对x进行尺取
{
    int s=0,t=0,sum=0,ans=0;
    while(t<2000){
        while(sum<x&&t<2000){
            sum+=p[t];
            t++;
        }
        if(sum<x)break;
        if(sum==x)ans++;
        sum-=p[s];
        s++;
    }
    return ans;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    getprime();
    for(int i=2;i<=10000;i++)ans[i]=solve(i);
    int n;
    while(cin>>n,n){
        cout<<ans[n]<<endl;
    }
    return 0;
}