Sum of Different Primes

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 expressn 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 thatn ≤ 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

#include<stdio.h>#include<string.h>/*内存使用比较大，状态设成3维的数组，dp[i][j][k]，取到第i个质数用了k个数凑成j的方案数*/int dp[200][1121][15],primes[200],isPrime[1121];int main(){    int i,j,k,n,cal;    memset(isPrime,1,sizeof(isPrime));    cal=0;    /*求出所有需要的素数*/    for(i=2;i<1121;i++){        if(isPrime[i]){            primes[cal++]=i;            for(j=2;j*i<1121;j++)                isPrime[i*j]=0;        }    }    /*dp----01背包*/    dp[0][0][0]=1;     for(i=1;i<=cal;i++){        for(j=0;j<1121;j++){            for(k=0;k<15;k++){                dp[i][j][k]=dp[i-1][j][k];                if(primes[i-1]<=j&&k)                     dp[i][j][k]+=dp[i-1][j-primes[i-1]][k-1];            }        }    }    while(scanf("%d %d",&n,&k)&&(n||k))        printf("%d\n",dp[cal][n][k]);    return 0;}

这是个01背包的问题，设计的状态是取到第I个质数用了k个数凑成j的方案数，上面的代码是三维数组，内存12560k,时间0ms下面进行对内存的优化...

下面的代码内存236KB，时间47ms...
#include<stdio.h>#include<string.h>/*内存使用比较小，状态设成2维的数组*/int dp[1121][15],primes[200],isPrime[1121];int main(){    int i,j,k,n,cal;    memset(isPrime,1,sizeof(isPrime));    cal=0;    for(i=2;i<1121;i++){        if(isPrime[i]){            primes[cal++]=i;            for(j=2;j*i<1121;j++)                isPrime[i*j]=0;        }    }    dp[0][0]=1;    for(i=1;i<=cal;i++){        for(j=1120;j>=0;j--){            for(k=14;k>=0;k--){                if(primes[i-1]<=j&&k)                    dp[j][k]+=dp[j-primes[i-1]][k-1];            }        }    }    while(scanf("%d %d",&n,&k)&&(n||k))        printf("%d\n",dp[n][k]);    return 0;}