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