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=,f=;char ch=getchar();
     while(ch<'' || ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>='' && ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 const int mxn=;
 int pri[mxn],cnt=;
 int vis[mxn];
 int n,k;
 int f[mxn][];
 void Pri(){
     int i,j;
     for(i=;i<=n;i++){
         if(!vis[i]){
             pri[++cnt]=i;
         }
         for(j=;j<=cnt && i*pri[j]<=n;j++){
             vis[i*pri[j]]=;
             if(i%pri[j]==)break;
         }
     }
     return;
 }
 int main(){
     n=;
     Pri();
     int i,j;
     f[][]=;
     for(i=;i<=cnt;i++){
         for(j=n;j>=pri[i];j--){
             for(int k=;k<=;k++)
                 f[j][k]+=f[j-pri[i]][k-];
         }
     }
     while(scanf("%d%d",&n,&k) && n){
         printf("%d\n",f[n][k]);
     }
     return ;
 }