POJ - 2739 Sum of Consecutive Prime Numbers_poj 2739sum of consecutive prime numbers-优快云博客

本文针对POJ2739问题提供了解决方案，该问题是关于如何使用一个或多个连续的素数之和来表示给定的正整数，并计算这种表示方式的数量。文章详细介绍了使用筛法查找素数的过程，并给出了完整的C语言实现代码。

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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：题意很简单，无非就是给出一个数用连续素数和表示的最多可能。遇0题目结束。

分析：首先看题目给你的数字，本身是不是素数；其次咱们从后往前找，直到找到2这个最小的质数，看看在这个过程中有几种可能；最后将结果输出。

那么直接上代码：

用筛法去查找素数，无疑是最简单快速的。

#include<stdio.h>
int b[10001]={0};
void num()
{
    b[0]=b[1]=1;
    for(int i=2;i<10001;i++)
    {
        if(b[i]==0)
        {
            for(int j=2*i;j<10001;j+=i) b[j]=1;
        }
    }
}
int main()
{
    num();
    int n;
    while(scanf("%d",&n)&&n)
    {
        int s=0;
        for(int i=n;i>=2;i--)
        {
            int ok=0;
            if(b[i]!=0) continue;
            for(int j=i;j>=2;j--)
            {
                if(b[j]==0&&j==n)
                {
                    s++;
                    break;
                }
                else
                {
                    if(b[j]==0) ok+=j;
                    if(ok==n)
                    {
                        s++;
                        break;
                    }
                    else if(ok>n) break;
                }
            }
        }
        printf("%d\n",s);
    }
    return 0;
}