解题报告_sum of consecutive prime numbers-优快云博客

本文链接：https://blog.youkuaiyun.com/shaomjc/article/details/81109755

该程序旨在确定正整数（2至10000）有多少种表示为连续素数之和的方式。输入序列包含多个正整数，以零结束。输出是每行对应输入的正整数的表示方式数量，素数必须连续。示例显示了不同整数的解决方案计数。

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Sum of Consecutive Prime Numbers
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 26954 Accepted: 14544

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

2 3 17 41 20 666 12 53 0

Sample Output

1 1 2 3 0 0 1 2

Sour

题目大意

给你一个数（1到1w），输出其是否能由连续的素数相加而成，若能输出总共的方案数；否则，输出0

题目思路

先用筛法生成一个素数表，然后对于每个输入的数直接进行枚举

#include<stdio.h>
#include<math.h>
int main()
{
    long long n,a[10000],q,s,p,w,i,k,j,h;
    for(;;)
    {
        w=0;
        q=0;
        s=0;
        scanf("%lld",&n);
        p=n;
        if(n==0)
            break;
        for(i=2; i<=n; i++)
        {
            k=sqrt(i);
            for(j=2; j<=k; j++)
                if(i%j==0)
                    break;
            if(j==k+1)
            {
                a[q++]=i;
                s+=1;
            }
        }
        for(i=s-1; i>=0; i--)
        {
            h=i;
            p=n;
            p=p-a[i];
            for(;;)
            {
                if(p<0)
                    break;
                else if(p!=0)
                {
                    if(h<=0)
                        break;
                    p=p-a[--h];
                }
                else if(p==0)
                {
                    w+=1;
                    break;
                }
            }
        }
        printf("%lld\n",w);
    }
    return 0;
}