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]={,},f[][N],t;
 int prime(){
     int t=,i,j,flag;
     for(i=;i<N;i+=){
         for(j=,flag=;prim[j]*prim[j]<=i;j++)
             if(i%prim[j]==) flag=;
         if(flag){
             prim[t++]=i;
         }
     }
     return t-;
 }
 void s(){
      for(int i=;i<=t;i++){
                 for(int j=;j>=;j--){
                     for(int p=;p>=prim[i];p--)
                         f[j][p]+=f[j-][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,,sizeof(f));
         f[][]=;
         if(k==&&n==) break;
         s();
          cout<<f[k][n]<<endl;
     }
     return ;
 }