POJ2739 Sum of Consecutive Prime Numbers

Sum of Consecutive Prime Numbers

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 25225 Accepted: 13757

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

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题目的意思是给出一个数n，求用连续的若干素数相加表示它，有多少种方法

思路：因为要求连续，而且相加有单调性，所以先打个素数表，再尺取法

#include <iostream>#include <cstring>#include <cstdio>#include <map>#include <queue>#include <cmath>#include <algorithm>#include <set>using namespace std;#define LL long longconst int inf=0x3f3f3f3f;int n,m;int a[10005];bool isp(int x){    if(x<2)        return 0;    for(int i=2; i<=sqrt(x); i++)    {        if(x%i==0)            return 0;    }    return 1;}int main(){    int n;    int cnt=0;    for(int i=1; i<10005; i++)    {        if(isp(i))            a[cnt++]=i;    }    while(~scanf("%d",&n)&&n)    {        int l=0,r=0,sum=0,cnt=0;        while(1)        {            while(a[r]<=n&&sum<=n)            {                sum+=a[r++];                if(sum==n)                    cnt++;            }            if(sum<=n) break;            sum-=a[l++];            if(sum==n) cnt++;        }        printf("%d\n",cnt);    }    return 0;}