POJ 2739 Sum of Consecutive Prime Numbers（线性素数筛+前缀和）

最新推荐文章于 2020-10-07 21:58:11 发布

原创最新推荐文章于 2020-10-07 21:58:11 发布 · 534 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#ACM #POJ #前缀和 #线性素数筛

POJ 专栏收录该内容

28 篇文章

订阅专栏

本文介绍了一种算法，用于解决特定整数可以由多少种连续素数之和表示的问题。通过线性素数筛算法预处理10000以内的所有素数并计算前缀和，再通过枚举起始和结束位置来统计每种可能的连续素数之和，最终输出给定正整数的所有可能表示方法的数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Sum of Consecutive Prime Numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 22735		Accepted: 12432

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

解题思路：先用线性素数筛算法O（n），算出10000以内1229个素数，并打表，然后算出这1229个素数的前缀和，而后分别枚举起点和终点，算出中间的连续素数和，然后让对应的数组+1.

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int b[10005];
int prime[2000];
int a[1300];
//10000以内1229个素数
int main()
{
    memset(b,0,sizeof(b));
    int i = 0,j;
    for(i=3;i<=10000;i+=2)//筛法求素数
    {
        if(!b[i])
        {
            for(j=i*2;j<=10000;j+=i)//将他的全部倍数排除
            {
                b[j] = 1;
            }
        }
    }
    j = 1;
    prime[0] = 2;
    for(i=3;i<=10000;i+=2)//将全部素数打表
    {
        if(!b[i])
        {
            prime[j++] = i;
        }
    }
    a[0] = 0;
    for(i=1;i<=1229;i++)//计算前缀和
    {
       a[i] = prime[i-1] + a[i-1];
       //cout << a[i] << endl;
    }
    memset(b,0,sizeof(b));
    for(i=0;i<=1227;i++)//枚举起点和终点
    {
        for(j=1228;j>i;j--)
        {
            if(a[j]-a[i]<=10000)
            b[a[j]-a[i]]++;
        }
    }
    int n;
    while(cin>>n && n)
    {
        cout << b[n] << endl;
    }
    //cout <<j << endl;
    //cout << "Hello world!" << endl;
    return 0;
}