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

题目链接：

POJ:

id=3132">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个不同样的素数，他们的和为n。求这种组合有多少种。

PS:

刚看到这题没有什么思路! 暴力和搜索是肯定不行的。后来发现能够用类似背包的方法！

代码例如以下：

#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;
}