| 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 ;
}