POJ 2739 素数打表-----水题

http://poj.org/problem?id=2739

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

一开始拿到题目没有思路，后来，想到打表，之前写了一次，错了，重新写。感觉自己不是很自信，打表怕超时，其实素数没多多少的。题目没有给范围，不要怕，做一下应该能出来的。打完表，判断下，小于的继续加，等于的就+1，大于直接下一个。

#include<iostream>
using namespace std;
int main()
{
int a[10000];
a[0] = 2;
int num = 1;
for (int i = 3; i < 10000; i++)
{
int flag = 0;
for (int j = 2; j < i; j++)
{
if (i%j == 0)
{
flag = 1;
break;
}
}
if (flag == 0)
{
a[num] = i;
num++;
}
}
int n;
while (cin >> n)
{
if (n == 0)
break;
int sum = 0,Num=0;
for (int i = 0; i < num; i++)
{
sum = a[i];
for (int j = i+1;; j++)
{
if (sum == n)
{
Num++;
break;
}
else if (sum>n)
{
break;
}
else if (sum < n)
{
sum += a[j];
}
}
}
cout << Num << endl;
}
}