POJ2739Sum of Consecutive Prime Numbers

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

分析：1、对每个输入n，计算出小于n的素数，然后从连续的1个数、2个数、。。。、n/2个数，穷举，超时；
　　　2、由于最大为10000，先建立素数表，然后从连续的1个数、2个数、。。。、n/2个数，穷举，超时；
　　　3、建立素数表，然后穷举结束条件为和大于等于数n，AC。

方法一：超时

#include <iostream>
#include <cmath>
#include <cstring>
using namespace std;

int prime[1229];

int main()
{
    int n, res;
    int i, j, length, end, start, flag;
    res=0;
    //memset(prime,0,sizeof(prime));

    for(i=2; i<=10000; i++)
    {
        int sq=sqrt(i);
        flag=0;
        for(j=2; j<=sq; j++)
        {
            if(i%j==0)
            {
                flag=1;
                break;
            }
        }
        if(flag==0)
        {
            prime[res]=i;
            res++;
        }
    }

    while(cin>>n&&n!=0)
    {
        int k;
        for(k=0;k<1229;k++){
            if(prime[k]>=n)
                break;
        }
        int sum=0, max=k;
        res=0;
        for(length=0; length<=k/2; length++)
        {
            start=0;
            end= max-length;

            for(start=0; start<=end; start++)
            {

                sum=0;
                for(i=start; i<=start+length; i++)
                {

                    sum=sum+prime[i];
                }
                if(sum==n)
                {
                    res++;
                    sum=0;
                    break;
                }
            }
        }
        cout<<res<<endl;
    }
    return 0;
}

方法二：AC

#include <iostream>
#include <cmath>
#include <cstring>
using namespace std;

int prime[1229];

int main()
{
    int n, res;
    int i, j, k, flag, sum;
    res=0;
    //memset(prime,0,sizeof(prime));

    for(i=2; i<=10000; i++)
    {
        int sq=sqrt(i);
        flag=0;
        for(j=2; j<=sq; j++)
        {
            if(i%j==0)
            {
                flag=1;
                break;
            }
        }
        if(flag==0)
        {
            prime[res]=i;
            res++;
        }
    }

    while(cin>>n&&n!=0)
    {
        k=0;
        for(i=0; i<res&&prime[i]<=n; i++)
        {
            sum = prime[i];
            j=i+1;
            while(sum<n&&j<res)
            {
                sum=sum+prime[j++];
            }
            if(sum==n)
            {
                k++;
            }
            //cout<<" i="<<i<<endl;
        }
        cout<<k<<endl;
    }
    return 0;
}