POJ 3132 Sum of Different Primes ( 满背包问题)

Sum of Different Primes

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 3280		Accepted: 2040

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 nas 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

题目大意：

　　这道题是说，给你一个数字ｎ，然后让你用仅仅含有ｋ个素数的集合来描述（就是取ｋ个素数使得他们的和是ｎ），问这样的集合有多少个．

解题思路：

　　一看到后，就想到了背包问题，把这个背包塞满的问题，定义如下状态：

　　dp[i][j]表示的是把数字i分成j个素数的集合的所有情况数。

　　初始状态：dp[1][1] = 0;//从题目中的信息知道的

　　　　　　　dp[0][0] = 1;//这个初值条件是硬加上去的，要不然的话dp[0][0]就无法递推了，永远都是0了。QAQ

代码：

# include<cstdio>
# include<iostream>
# include<fstream>
# include<algorithm>
# include<functional>
# include<cstring>
# include<string>
# include<cstdlib>
# include<iomanip>
# include<numeric>
# include<cctype>
# include<cmath>
# include<ctime>
# include<queue>
# include<stack>
# include<list>
# include<set>
# include<map>

using namespace std;

const double PI=4.0*atan(1.0);

typedef long long LL;
typedef unsigned long long ULL;

# define inf 999999999
# define MAX 1120+4


int dp[MAX][MAX];//dp[i][j]表示把数字i分解为由j个素数组成的集合数目
int book[MAX];
int prime[MAX];
int len;


void init()
{
    for ( int i = 2;i <= MAX;i++ )
    {
        book[i] = 1;
    }
    for ( int i = 2;i <= MAX;i++ )
    {
        if ( book[i]==1 )
        {
            for ( int j = 2*i;j <= MAX;j+=i )
            {
                book[j] = 0;
            }
        }
    }
    len = 0;

    for ( int i = 2;i <= MAX;i++ )
    {
        if ( book[i]==1 )
        {
            prime[++len] = i;
        }
    }
}

void solve()
{
        memset(dp,0,sizeof(dp));
        dp[0][0] = 1;
        dp[1][1] = 0;
        for ( int i = 1;i <= len;i++ )
        {
            for ( int j = MAX;j >= prime[i];j-- )
            {
                for ( int k = 1;k <= 14;k++ )
                {
                    dp[j][k] += dp[j-prime[i]][k-1];
                }
            }
        }
}


int main(void)
{

    init();
    int n,k;
    while ( cin>>n>>k )
    {
        if ( n==0&&k==0 )
            break;
        solve();
        cout<<dp[n][k]<<endl;
    }


    return 0;
}