POJ 3132 Sum of Different Primes

题目大意：输入n，k，要求你找出k个和为n的不相同的素数，求可行的方案数。。。

 

Description

A positive integer may be expressed as a sum of different prime numbers (primes), in one way or another. Given two positive integersn andk, you should count the number of ways to express n as a sum ofk 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. Forn = 24 andk = 2, the answer is three, because there are three sets {5, 19}, {7, 17} and {11, 13}. Forn = 2 andk = 1, the answer is one, because there is only one set {2} whose sum is 2. Forn = 1 andk = 1, the answer is zero. As 1 is not a prime, you shouldn’t count {1}. Forn = 4 andk = 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 givenn andk.

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 integersn andk separated by a space. You may assume that n ≤ 1120 andk ≤ 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 forn andk 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

/*一开始想用搜索做来着，可惜没成功。。。。。。*/#include<stdio.h>#define M 1125int prime[M]={2,3,5,7,11};__int64 dp[M][50];int num=5;int judge(int n){int i,j;for(i=0;prime[i]*prime[i]<=n;i++){if(n%prime[i]==0) return 0;}return 1;}void Init(){int gab,i,j;for(gab=2,i=13;i<M;i+=gab){gab=6-gab;if(judge(i)) {//printf("%d==%d\n",num,i);prime[num++]=i;}}}int main(){int i,j,k;int sum,n;Init();dp[0][0]=1;  /*习惯性地都会输入一个处理一个。。嗯嗯。。借鉴了别人的~很不错的提前处理~~~*/for(i=0;i<num;i++){  /*if写成：(i=0;i<=num;i++)  就 WA了。。额。。太低级的错误了！！！因为num是个数，prime[num]==0啊！！！！！！*/for(j=M;j>=prime[i];j--){for(k=15;k>0;k--){dp[j][k]+=dp[j-prime[i]][k-1];}}}while(scanf("%d%d",&sum,&n),sum+n){printf("%I64d\n",dp[sum][n]);}return 0;}