Sum of Consecutive Prime Numbers----暴力水题

Sum of Consecutive Prime Numbers

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 24890 Accepted: 13552

Description

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

2317412066612530

Sample Output

11230012

题目链接：http://poj.org/problem?id=2739

题目的意思就是给你一个数，有几种不同的素数组合方案会组成这个数。

本以为是个数论难题，但是写了发暴力，竟然就过了。。。过了。。

先素数筛打表，然后暴力即可

代码：

#include <cstdio>#include <cstring>#include <iostream>#include <cmath>using namespace std;bool vis[20000];int prime[3000];int top;void Init(){    memset(vis,false,sizeof(vis));    for(int i=2;i<=10000;i++){        if(vis[i])            continue;        prime[top++]=i;        for(int j=i*2;j<=10000;j+=i){            vis[j]=true;        }    }}int main(){    top=0;    Init();    int t,sum,k;    while(1){        scanf("%d",&t);        if(t==0)            break;        k=0;        for(int i=0;i<top;i++){            sum=0;            for(int j=i;j<top;j++){                sum+=prime[j];                if(sum==t){                    k++;                    break;                }                else if(sum>t){                    break;                }            }        }        cout<<k<<endl;    }}