poj 3132

Sum of Different Primes

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 3360		Accepted: 2092

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

思路：prim[]为素数表；
　　　f[i][j]为j拆分成i个素数和的方案数（1<=i&&i<=14,prim[i]<=j&&j<=1199) 边界f[0][0]=1;
　　　int num 为prim[]的表长；
　　　使用DP计算k个不同素数的和为n的方案总数：
　　　　　　枚举prim[]中的prim[i](0<=i&&i<=num);
　　　　　　　　按递减顺序枚举素数的个数j(14>=j&&j>=1);
　　　　　　　　　　　递减枚举前j个素数的和p(1199>=p&&p>=prim[i]);
　　　　　　　　　　　　　　累计prim[i]作为第j个素数的方案总数f[j][p]+=f[j-1][p-prim[i]];
　　　　　　
　　　f[k][n]即为解！！！！！！

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<iomanip>
#include<cmath>
#include<vector>
#include<queue>
#include<stack>
using namespace std;
#define PI 3.141592653589792128462643383279502
#define N 1200
int prim[N]={2,3},f[15][N],t;
int prime(){
    int t=2,i,j,flag;
    for(i=5;i<N;i+=2){
        for(j=0,flag=1;prim[j]*prim[j]<=i;j++)
            if(i%prim[j]==0) flag=0;
        if(flag){
            prim[t++]=i;
        }
    }
    return t-1;
}
void s(){
     for(int i=0;i<=t;i++){
                for(int j=14;j>=1;j--){
                    for(int p=1199;p>=prim[i];p--)
                        f[j][p]+=f[j-1][p-prim[i]];
                }
            }
}
int main(){
    //#ifdef CDZSC_June
    //freopen("in.txt","r",stdin);
    //#endif
    //std::ios::sync_with_stdio(false);
    t=prime();
    int k,n;
    while(scanf("%d%d",&n,&k)){
        memset(f,0,sizeof(f));
        f[0][0]=1;
        if(k==0&&n==0) break;
        s();
         cout<<f[k][n]<<endl;
    }
    return 0;
}