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

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 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：

 1 # include<cstdio>
 2 # include<iostream>
 3 # include<fstream>
 4 # include<algorithm>
 5 # include<functional>
 6 # include<cstring>
 7 # include<string>
 8 # include<cstdlib>
 9 # include<iomanip>
10 # include<numeric>
11 # include<cctype>
12 # include<cmath>
13 # include<ctime>
14 # include<queue>
15 # include<stack>
16 # include<list>
17 # include<set>
18 # include<map>
19
20 using namespace std;
21
22 const double PI=4.0*atan(1.0);
23
24 typedef long long LL;
25 typedef unsigned long long ULL;
26
27 # define inf 999999999
28 # define MAX 1120+4
29
30
31 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
32 int book[MAX];
33 int prime[MAX];
34 int len;
35
36
37 void init()
38 {
39     for ( int i = 2;i <= MAX;i++ )
40     {
41         book[i] = 1;
42     }
43     for ( int i = 2;i <= MAX;i++ )
44     {
45         if ( book[i]==1 )
46         {
47             for ( int j = 2*i;j <= MAX;j+=i )
48             {
49                 book[j] = 0;
50             }
51         }
52     }
53     len = 0;
54
55     for ( int i = 2;i <= MAX;i++ )
56     {
57         if ( book[i]==1 )
58         {
59             prime[++len] = i;
60         }
61     }
62 }
63
64 void solve()
65 {
66         memset(dp,0,sizeof(dp));
67         dp[0][0] = 1;
68         dp[1][1] = 0;
69         for ( int i = 1;i <= len;i++ )
70         {
71             for ( int j = MAX;j >= prime[i];j-- )
72             {
73                 for ( int k = 1;k <= 14;k++ )
74                 {
75                     dp[j][k] += dp[j-prime[i]][k-1];
76                 }
77             }
78         }
79 }
80
81
82 int main(void)
83 {
84
85     init();
86     int n,k;
87     while ( cin>>n>>k )
88     {
89         if ( n==0&&k==0 )
90             break;
91         solve();
92         cout<<dp[n][k]<<endl;
93     }
94
95
96     return 0;
97 }