Sum of Different Primes
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 3360 | Accepted: 2092 |
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 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 思路：prim[]为素数表； f[i][j]为j拆分成i个素数和的方案数（1<=i&&i<=14,prim[i]<=j&&j<=1199) 边界f[0][0]=1; int num 为prim[]的表长； 使用DP计算k个不同素数的和为n的方案总数： 枚举prim[]中的prim[i](0<=i&&i<=num); 按递减顺序枚举素数的个数j(14>=j&&j>=1); 递减枚举前j个素数的和p(1199>=p&&p>=prim[i]); 累计prim[i]作为第j个素数的方案总数f[j][p]+=f[j-1][p-prim[i]]; f[k][n]即为解！！！！！！
1 #include<iostream>
2 #include<cstdio>
3 #include<algorithm>
4 #include<cstring>
5 #include<cstdlib>
6 #include<iomanip>
7 #include<cmath>
8 #include<vector>
9 #include<queue>
10 #include<stack>
11 using namespace std;
12 #define PI 3.141592653589792128462643383279502
13 #define N 1200
14 int prim[N]={2,3},f[15][N],t;
15 int prime(){
16 int t=2,i,j,flag;
17 for(i=5;i<N;i+=2){
18 for(j=0,flag=1;prim[j]*prim[j]<=i;j++)
19 if(i%prim[j]==0) flag=0;
20 if(flag){
21 prim[t++]=i;
22 }
23 }
24 return t-1;
25 }
26 void s(){
27 for(int i=0;i<=t;i++){
28 for(int j=14;j>=1;j--){
29 for(int p=1199;p>=prim[i];p--)
30 f[j][p]+=f[j-1][p-prim[i]];
31 }
32 }
33 }
34 int main(){
35 //#ifdef CDZSC_June
36 //freopen("in.txt","r",stdin);
37 //#endif
38 //std::ios::sync_with_stdio(false);
39 t=prime();
40 int k,n;
41 while(scanf("%d%d",&n,&k)){
42 memset(f,0,sizeof(f));
43 f[0][0]=1;
44 if(k==0&&n==0) break;
45 s();
46 cout<<f[k][n]<<endl;
47 }
48 return 0;
49 }