POJ 2739 Sum of Consecutive Prime Numbers

Sum of Consecutive Prime Numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 25167		Accepted: 13722

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
题目大意：
给定一个10000以内的数字，判断这个数字是否可以由几个连续的素数（例如：2,3,5,7...）相加得到，并且给出这个数可以有几组这样的解。
比如41 = 2+3+5+7+11+13， 41 = 11+13+17， 41 = 41
输入：每行一个数字，0为退出
输出：每行一个数字，对应输入的每个数字的解的组数。
解题思路：
可以优化解题思路，减少时间。10000以内素数总共有1229个。可以先打表将所有素数存到数组中，然后再设置两个变量i，j，分别指向这个连续素数串的头和尾，不停的移动寻找解即可。下面是AC代码：
#include<iostream>
#include<cmath>
using namespace std;
int is_prime(int num )
{
    int i;
    if(num==2)
        return 1;
    for(i=2; i<(int)sqrt((double)num); i++)
    {
        if(num%i==0)
        break;
    }
    if(num%i==0)
        return 0;
    else return 1;
}
int main()
{
    ios::sync_with_stdio(false);
    int i,j=0;
    int n;
    int a[10000];
    for(i=2; i<10000; i++)
    {
        if(is_prime(i)==1)
            a[j++]=i;
    }
    while(cin>>n&&n)
    {
        int num=0;
        for(i=0; i<2000; i++)
        {
            int sum=0;
            for(j=i; j<2000; j++)
            {
                sum+=a[j];
                if(sum>n)
                    break;
                else if(sum==n)
                {
                    num++;
                    break;
                }
            }
        }
        cout<<num<<endl;
    }
    return 0;
}