POJ 2739 Sum of Consecutive Prime Numbers( *【素数存表】+暴力枚举 )

Sum of Consecutive Prime Numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 19895		Accepted: 10906

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
 题目分析:给你一个数，如果这个数能够表示成几个连续的素数相加的形式，请问有多少种这样的形式。
 例如：41=2+3+5+7+11+13
      41=11+13+17
      41=41   共有3种
像这种7+13=20 或者 3+5+5+7=20  这些都是不合法的。
算法分析：数据范围不大，先将1000以内的素数存在一个数组里，两层循环进行暴力枚举
        外层循环控制最小的素数累加和的起点数，从它开始一直累加下去直到>=输入数。
        如果累加和==n，计数器++，最后输出结果。
代码如下：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <string>
#include <algorithm>

using namespace std;

int prime[2000], total=0;

bool isprime(int k )
{
    for(int i=0; i<total; i++)
    {
        if(k%prime[i] == 0)
        {
            return false;
        }
    }
    return true;
}

int main()
{
    int i, j;
    for( i=2; i<=10000; i++)
    {
        if(isprime(i))
        {
            prime[total++]=i;
        }
    }
    prime[total]=10001;  //???

    int n;
    while(scanf("%d", &n)!=EOF)
    {
        if(n==0)
            break;
        int ans=0;
        for(i=0; n>=prime[i]; i++)
        {
            int cnt=0;
            for(j=i; j<total &&cnt<n; j++)
            {
                cnt+=prime[j];
            }
            if(cnt==n)
            {
                ++ans;
            }
        }
        printf("%d\n", ans );
    }
    return 0;
}