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;

}