poj 2739 Sum of Consecutive Prime Numbers（尺取法）

最新推荐文章于 2020-10-07 21:58:11 发布

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本文探讨了如何找出一个正整数可以表示为连续质数之和的不同方式的数量。通过预处理所有小于等于10000的质数，并使用滑动窗口的方法来寻找所有可能的连续质数序列，使得这些序列的和等于给定的正整数。

Sum of Consecutive Prime Numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 22876		Accepted: 12509

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

Sample Output

题意：每一个数都可以由连续的质数相加得到，给一个数n，问一共有多少个质数序列和为n

思路：先存下所有的质数，因为质数是有序的，我们就可以用尺取法去枚举了。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 11111
int a[N],vis[N];
int cnt;
void init()
{
    cnt=0;
    memset(vis,0,sizeof(vis));
    for(int i=2;i<=10000;i++)
    {
        if(vis[i]) continue;
        a[cnt++]=i;
        for(int j=i*i;j<=10000;j+=i)
            vis[j]=1;
    }
}
int main()
{
    init();
    int n;
    while(~scanf("%d",&n)&&n)
    {
        int l=0,r=0;
        int mat=0,ans=0;
        for(int i=0;i<cnt;i++)
        {
            if(a[i]>n) break;
            while(r<cnt&&mat<n)
                {
                    mat+=a[r];
                    r++;
                }
            if(mat==n)  ans++;
            mat-=a[l++];
        }
        printf("%d\n",ans);
    }
    return 0;
}