C - Sum of Consecutive Prime Numbers(1.2.2)

C - Sum of Consecutive Prime Numbers(1.2.2)

Time Limit:1000MS    Memory Limit:65536KB    64bit IO Format:%I64d & %I64u

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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
#include <iostream>
#include <vector>
using namespace std;
vector <int> prime;
int len=0;
bool is_prime(int n)
{
    int i;
    if(n<=1)    return false;
    else if(n==2)   return true;
    else if(n%2==0) return false;
    else
        for(i=3;i*i<=n;i+=2)
            if(n%i==0)  return false;
    return true;
}
void primes()
{
    int i;
    prime.push_back(0);          //将第一个数初始化为0，以便用于求输入的数本身是否为素数
    for(i=2;i<=10000;i++)
        if(is_prime(i))
            prime.push_back(i);  //找出10000以内所有的素数
    len = prime.size();
    for(i=1;i<len;i++)
    {
        prime[i] += prime[i-1]; //利用动态规划，计算出每个素数与其之前所有素数之和
    }
}
int main()
{
    int Posinteger,count,i,j;
    primes();
    while(cin>>Posinteger&&Posinteger)
    {
        count = 0;
        for(i=0;i<len;i++)
        {
            for(j=i+1;j<len;j++)
                if(prime[j]-prime[i]==Posinteger)   //判断前j个素数之和与前i个素数之和的差是否与Posinteger相等，
                    count++;                       //如果相等，则表明有一组连续的素数相加之和与之相等，符合题意。
        }
        cout<<count<<endl;
    }

    return 0;
}