POJ3132 Sum of Different Primes

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 3473		Accepted: 2154

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

Source

Japan 2006

 

用素数筛打一个素数表出来，然后在素数表上背包动规。

/*by SilverN*/
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
const int mxn=5000;
int pri[mxn],cnt=0;
int vis[mxn];
int n,k;
int f[mxn][30];
void Pri(){
    int i,j;
    for(i=2;i<=n;i++){
        if(!vis[i]){
            pri[++cnt]=i;
        }
        for(j=1;j<=cnt && i*pri[j]<=n;j++){
            vis[i*pri[j]]=1;
            if(i%pri[j]==0)break;
        }
    }
    return;
}
int main(){
    n=1250;
    Pri();
    int i,j;
    f[0][0]=1;
    for(i=1;i<=cnt;i++){
        for(j=n;j>=pri[i];j--){
            for(int k=1;k<=15;k++)
                f[j][k]+=f[j-pri[i]][k-1];
        }
    }
    while(scanf("%d%d",&n,&k) && n){
        printf("%d
",f[n][k]);
    }
    return 0;
}