UVa 1213 Sum of Different Primes (DP)

题意：给定两个数 n 和 k，问你用 k 个不同的质数组成 n，有多少方法。

析：dp[i][j] 表示 n 由 j 个不同的质数组成，然后先打表素数，然后就easy了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define print(a) printf("%d
", (a))
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1120 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}

int a[maxn];
vector<int> prime;
int dp[maxn][15];

int main(){
    memset(a, 0, sizeof(a));
    m = (int)sqrt(maxn + 0.5);
    for(int i = 2; i <= m; i++)   if(!a[i])
        for(int j = i * i; j < maxn; j += i)    a[j] = 1;
    for(int i = 2; i < maxn; ++i)  if(!a[i])  prime.push_back(i);
    memset(dp, 0, sizeof dp);

    dp[0][0] = 1;
    for(int k = 0; k < prime.size(); ++k){
        for(int i = 1120; i >= prime[k]; --i){
            for(int j = 1; j <= 14; ++j)
                dp[i][j] += dp[i-prime[k]][j-1];
        }
    }

    while(scanf("%d %d", &n, &m) == 2 && m+n)  printf("%d
", dp[n][m]);
    return 0;
}