POJ3132Sum of Different Primes题解动态规划DP

Sum of Different Primes

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 2160		Accepted: 1334

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

简单的二维费用01背包

状态：

d[i][j]表示j个素数组成和为i的方法数

状态转移方程：

d[k][j]+=d[k-p[i]][j-1];

边界：

d[0][0]=1;

代码：

#include<cstdio> #define N 1200 int p[N]; bool f[N]; void init() { int i,j,m=0; f[0]=f[1]=1; for(i=4;i<=N;i+=2) f[i]=1; for(i=3;i*i<=N;i+=2) if(!f[i]) for(j=i*i;j<=N;j+=i*2) f[j]=1; for(i=2;i<=N;i++) if(!f[i]) p[m++]=i; } int main() { init(); int n,m; while(scanf("%d%d",&n,&m),n+m) { int i,j,k,d[N][15]={0}; d[0][0]=1; for(i=0;p[i]<=n;i++) for(k=n;k>=p[i];k--) for(j=m;j>0;j--) d[k][j]+=d[k-p[i]][j-1]; printf("%d/n",d[n][m]); } }