POJ 3132 Sum of Different Primes

Description
A positive integer may be expressed as a sum of different prime numbers (primes), in one way or another. Given two positive integers n and k, you should count the number of ways to express n as a sum of k different primes. Here, two ways are considered to be the same if they sum up the same set of the primes. For example, 8 can be expressed as 3 + 5 and 5 + 3 but the are not distinguished.
When n and k are 24 and 3 respectively, the answer is two because there are two sets {2, 3, 19} and {2, 5, 17} whose sums are equal to 24. There are not other sets of three primes that sum up to 24. For n = 24 and k = 2, the answer is three, because there are three sets {5, 19}, {7, 17} and {11, 13}. For n = 2 and k = 1, the answer is one, because there is only one set {2} whose sum is 2. For n = 1 and k = 1, the answer is zero. As 1 is not a prime, you shouldn’t count {1}. For n = 4 and k = 2, the answer is zero, because there are no sets of two different primes whose sums are 4.
Your job is to write a program that reports the number of such ways for the given n and k.
Input
The input is a sequence of datasets followed by a line containing two zeros separated by a space. A dataset is a line containing two positive integers n and k separated by a space. You may assume that n ≤ 1120 and k ≤ 14.
Output
The output should be composed of lines, each corresponding to an input dataset. An output line should contain one non-negative integer indicating the number of the ways for n and k specified in the corresponding dataset. You may assume that it is less than 231.
Sample Input
24 3
24 2
2 1
1 1
4 2
18 3
17 1
17 3
17 4
100 5
1000 10
1120 14
0 0
Sample Output
2
3
1
0
0
2
1
0
1
55
200102899
2079324314

---

先用线性筛，筛素数，然后就是01背包了。

---

#include<iostream>
#include<cstdio>
using namespace std;
const int maxn=1200;
int cnt,n,k,pri[maxn],dp[maxn][21];
bool vis[maxn];
void euler(int m)
{
    vis[1]=1;
    for(int i=2;i<=m;i++)
    {
        if(!vis[i])
            pri[++cnt]=i;
        for(int j=1;i*pri[j]<=m&&j<=cnt;j++)
        {
            vis[i*pri[j]]=1;
            if(i%pri[j]==0)
                break;
        }
    }
}
int main()
{
    euler(1120);
    dp[0][0]=1;
    for(int i=1;i<=cnt;i++)
        for(int j=1120;j>=pri[i];j--)
            for(int k=1;k<=15;k++)
                dp[j][k]+=dp[j-pri[i]][k-1];
    while(scanf("%d%d",&n,&k)&&n&&k)
        printf("%d\n",dp[n][k]);
    return 0;
}