POJ 2739 Sum of Consecutive Prime Numbers_int a[10010]={0}-CSDN博客

本文链接：https://blog.csdn.net/hqd_acm/article/details/5901846

本文探讨了如何找出正整数能表示为一个或多个连续质数之和的方法及其实现。通过对质数进行预处理并使用高效算法，文章提供了一个程序解决方案来计算特定正整数的这种表示方式的数量。

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Sum of Consecutive Prime Numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9435		Accepted: 5407

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

Source

Japan 2005

#include<cstdio>
bool isprime[10010]={false};
int num=0,prime[5000];
int a[10010]={0};
void init()
{
   isprime[1]=isprime[0]=true;
   for(int i=2;i*i<=10010;i++)
   if(!isprime[i])
   {
       for(int j=i*i;j<=10010;j+=i)
       isprime[j]=true;
   }
   for(int i=2;i<=10010;i++)
   if(!isprime[i]) prime[++num]=i;
}
int main()
{
    init();
    for(int i=1;i<=10010;i++)
    {
        for(int j=1;j<=num;j++)
        {
            if(prime[j]==i)
            {
                a[i]++;
                break;
            }
            if(prime[j]>i) break;
            int sum=0;
            int t=j;
            while(sum<i)
            {
                sum+=prime[t];
                t++;
            }
            if(sum==i) a[i]++;
        }
    }
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0) break;
        printf("%d/n",a[n]);
    }
    return 0;
}