1716. Sum of Different Primes

单点时限: 5.0 sec

内存限制: 256 MB

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 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.

输入格式

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.

输出格式

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 2^31.

样例

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

output

2
3
1
0
0
2
1
0
1
55
200102899
2079324314

复盘： 0-1背包问题。

1、物品的遍历一定在最外层，否则会导致重复使用同一件物品的问题

2、容积与物品数都必须逆序遍历，否则也会出现重复使用同一件物品的问题

3、状态转移通常通过新出现一件物品来更新所有状态。

#include<iostream>
#include<vector>
#include<map>
using namespace std;

void initPrime(int x,map<int,int> &prime){
	vector<int> mark(x+1,1);
	for(int i=2;i<=x;i++){
		if(mark[i]){
			prime[i]=1;
			for(int j=i+i;j<=x;j+=i)mark[j]=0;
		}
	}
}

int main(){
	int maxN=1120,k=14,n,m;
	map<int,int> prime;
	initPrime(maxN,prime);
	vector<vector<int>> ans(k+1,vector<int>(maxN+1,0));
	ans[0][0]=1;
	for(auto&[a,b]:prime){
		for(int i=maxN;i>0;i--){
			for(int j=k;j>=0;j--){
				if(a<=i&&j>0)ans[j][i]+=ans[j-1][i-a];
			}
		}
	}
	while(scanf("%d %d",&n,&m)&&(n||m))
        printf("%d\n",ans[m][n]);
}