UVA 1213 Sum of Different Primes （不同素数之和）（dp）

题意：选择k个质数，使它们的和等于n，问有多少种方案。

分析：dp[i][j]，选择j个质数，使它们的和等于i的方法数。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b) {
    if(fabs(a - b) < eps)  return 0;
    return a < b ? -1 : 1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 1120 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int dp[MAXN][20];
int vis[MAXN];
void init(){
    vis[0] = vis[1] = 1;
    for(int i = 2; i <= sqrt(MAXN + 0.5); ++i){
        if(!vis[i]){
            for(int j = i * i; j < MAXN; j += i){
                vis[j] = 1;
            }
        }
    }
    dp[0][0] = 1;
    for(int i = 0; i < MAXN; ++i){
        if(vis[i]) continue;
        for(int j = 14; j >= 1; --j){
            for(int k = MAXN - 1; k >= i; --k){
                dp[k][j] += dp[k - i][j - 1];
            }
        }
    }
}
int main(){
    init();
    int n, k;
    while(scanf("%d%d", &n, &k) == 2){
        if(!n && !k) return 0;
        printf("%d\n", dp[n][k]);
    }
    return 0;
}