hdu 5378 概率dp 逆元

标签：acm   概率dp   hdu   

然后，double肯定是不能过的。于是傻傻用结构体存了个分数。然后分母乘起来爆long long了。然后和GT，喵呜三个debug了一下午 =。= 怪我不懂逆元，坑了两位。

逆元：http://cn.bing.com/ 大致就是除个数取模等于成该数的逆元再取模吧...

所以转移变成了:

然后dp就是在整数里面转移了。

/***********************************************
 ** problem ID    : hdu_5378.cpp
 ** create time    : Wed Aug 12 11:14:30 2015
 ** auther name    : xuelanghu
 ** auther blog    : blog.csdn.net/xuelanghu407
 **********************************************/

#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int MAXN = 1000 + 5;
const long long MOD  = (long long)(1e9 + 7);

struct Node {
    long long a, b;
    Node() {a = 0; b = 1;}
    Node(long long _a, long long _b): a(_a), b(_b) {}
};

long long _gcd(long long a, long long b) {
    if (b == 0) return a;
    else return _gcd(b, a%b);
}

inline Node add (Node x, Node y) {
    long long rec, res, tmp;
    rec = x.a*y.b + x.b*y.a;
    res = x.b*y.b;
    tmp = _gcd(rec, res);// if (tmp == 0) tmp = 1;
    return Node(rec/tmp, res/tmp);
}

inline Node mul (Node x, Node y) {
    long long rec, res, tmp;
    rec = x.a * y.a;
    res = x.b * y.b;
    tmp = _gcd(rec, res);// if (tmp == 0) tmp = 1;
    return Node(rec/tmp, res/tmp);
}

int n, k;
vector<int>  v[MAXN];
long long    dp[MAXN][MAXN];
long long    c[MAXN];


long long dfs(int t, int fa) {
    long long cnt = 1L;
    for (int i=0; i<v[t].size(); i++) {
        if (v[t][i] == fa) continue;
        cnt += dfs(v[t][i], t);
    }
    return c[t] = cnt;
} 

long long power(long long a, long long p) {  
    long long res=1L;  
    while(p) {  
        if(p&1) res = (res*a)%MOD;  
        a = (a*a)%MOD;  
        p >>= 1;  
    }  
    return res;  
}  
  
long long Inv(long long a) {  
    return power(a, MOD-2);  
}  

long long _In[MAXN];
void init() {
    for (int i=1; i<=1002; i++) {
        _In[i] = Inv(i);
    }
}

long long f(int n) {
    long long res = 1;
    for (int i=1; i<=n; i++) {
        res = (res * (long long) (i)) % MOD;
    }
    return res;
}

int main () {
    int T_case;
    init();
    scanf("%d", &T_case);
    for (int i_case=1; i_case<=T_case; ++i_case) {
        
        scanf ("%d%d", &n, &k);
        
        for (int i=0; i<=n; i++) v[i].clear();
        // memset(c, 0,sizeof(c));
        int a, b;
        for (int i=1; i<n; i++) {
            scanf ("%d%d", &a, &b);
            v[a].push_back(b);
            v[b].push_back(a);
        }
        
        dfs(1, -1);
        
        dp[1][0] = ((c[1]-1) * _In[c[1]])%MOD;
        dp[1][1] = 1 * _In[c[1]];
        
        // cout << dp[1][0].a << "/" << dp[1][0].b << "  ";
        // cout << dp[1][1].a << "/" << dp[1][1].b << endl;
        for (int i=2; i<=n; i++) {
            // dp[i][0] =  mul(dp[i-1][0], Node(c[i]-1, c[i]));
            dp[i][0] = ((dp[i-1][0] * (c[i]-1)) % MOD * _In[c[i]]) % MOD;
            
            for (int j=1; j<=min(i, k); j++) {
            // dp[i][j] = add(mul(dp[i-1][j], Node(c[i]-1, c[i])), mul(dp[i-1][j-1], Node(1, c[i])));
            
            dp[i][j] = ((((dp[i-1][j] * (c[i]-1)) % MOD) * _In[c[i]]) % MOD + (dp[i-1][j-1] * _In[c[i]]) % MOD) % MOD;
            
            // if (i == n && j == k) cout << "hahahaha~" << endl;
            // cout << dp[i][j].a << "/" << dp[i][j].b << "  ";
            // if(dp[i][j].b <= 0)  dp[n+100][k+1].b = 100;
            }// cout << endl;
        }
        
        long long res = dp[n][k];
        // cout << res.a << "/" << res.b << endl;
        long long ans = (f(n) * res) % MOD;
        
        printf ("Case #%d: %d\n", i_case, (int)ans); 
    }
    return 0;
}

hdu 5378 概率dp 逆元

标签：acm   概率dp   hdu   

原文地址：http://blog.csdn.net/xuelanghu407/article/details/47627475