| 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 }
原文：http://www.cnblogs.com/yoyo-sincerely/p/5074749.html