Sum of Consecutive Prime Numbers(POJ 2739)尺取法+数学问题

来自《挑战程序设计竞赛》

1.题目原文

http://poj.org/problem?id=2739

Sum of Consecutive Prime Numbers

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 23592 Accepted: 12887

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

Source

Japan 2005

2.解题思路

将一个整数分成连续素数之和，一共有多少种方法？

很符合尺取法的情况。很明显要先处理出不超过n的所有素数，采用埃式筛法。

3.AC代码

#include<algorithm>#include<cctype>#include<cmath>#include<cstdio>#include<cstdlib>#include<cstring>#include<iomanip>#include<iostream>#include<map>#include<queue>#include<string>#include<set>#include<vector>#include<cmath>#include<bitset>#include<stack>#include<sstream>#include<deque>#include<utility>using namespace std;#define INF 0x7fffffff#define maxn 10005int prime[maxn];bool is_prime[maxn];//is_prime[i]是true表示i是素数//返回n以内素数的个数int sieve(int n){    int p=0;    for(int i=0;i<=n;i++) is_prime[i]=true;    is_prime[0]=is_prime[1]=false;    for(int i=2;i<=n;i++){        if(is_prime[i]){            prime[p++]=i;            for(int j=2*i;j<=n;j+=i) is_prime[j]=false;        }    }    return p;}int n;void solve(){    int p=sieve(n);    int s=0,t=0,sum=0;    int res=0;    for(;;){       while(t<p&&sum<n){           sum+=prime[t++];       }       if(sum<n) break;       else if(sum==n){            res++;       }       sum-=prime[s++];    }    printf("%d\n",res);}int main(){    while(scanf("%d",&n)!=EOF&&n){        solve();    }    return 0;}