POJ 3132 & ZOJ 2822 Sum of Different Primes(dp)

题目链接：

POJ:http://poj.org/problem?id=3132
ZOJ:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2822

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 kdifferent 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

题意：

给出n和k，求找出k个不相同的素数，他们的和为n，求这样的组合有多少种。

PS:

刚看到这题没有什么思路! 暴力和搜索是肯定不行的，后来发现可以用类似背包的方法！

代码如下：

#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#include <iostream>using namespace std;const int maxn = 1700;int dp[maxn][17];int prime(int i){    int m = sqrt(i*1.0);    int j;    for(j = 2; j <= m; j++)    {        if(i%j == 0)            break;    }    if(j > m)        return 1;    return 0;}int p[maxn];int l = 0;void init(){    for(int i = 2; i <= maxn; i++)    {        if(prime(i))        {            p[l++] = i;        }    }}int main(){    int n, k;    init();    while(~scanf("%d%d",&n,&k))    {        if(n == 0 && k == 0)            break;        memset(dp,0,sizeof(dp));        dp[0][0] = 1;        for(int i = 0; i < l; i++)        {            for(int j = n; j >= p[i]; j--)            {                for(int x = k; x > 0; x--)                {                    dp[j][x]+=dp[j-p[i]][x-1];                }            }        }        printf("%d\n",dp[n][k]);    }    return 0;}