HDU 4958 DP

DP算法求解问题

最新推荐文章于 2019-07-22 09:55:53 发布

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//
//  main.cpp
//  HDU 4958 DP
//
//  Created by 郑喆君 on 8/18/14.
//  Copyright (c) 2014 itcast. All rights reserved.
//

#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>
#include <cassert>
#include <complex>
using namespace std;
typedef long long ll;
typedef long double ld;
const int int_max = 0x07777777;
const int int_min = 0x80000000;
const int inf=0x20202020;
const ll mod=1000000007;
const double eps=1e-9;
const double pi=3.1415926535897932384626;
const int DX[]={1,0,-1,0},DY[]={0,1,0,-1};
ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll powmod(ll a,ll b,ll mod) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
bool cmp (const void *a , const void *b )
{
    return *(int *)a < *(int *)b;
}
ld dp[1005][1005];
ld fn (int a, int b) {if(a<0||b<0) return 0; else return dp[a][b];}
int main(int argc, const char * argv[])
{
    dp[0][0] = 0;
    for(int i = 0; i < 1005; i++){
        for(int j = 0; j < 1005-i; j++){
            if(i||j){
                if(i%2==0){
                    dp[i][j] = i+j+fn(i-1,j)*i/(i+j) + fn(i,j-1)*j/(i+j);
                }else{
                    dp[i][j] = max(fn(i-1,j),fn(i, j-1));
                }
            }
        }
    }
    int t;
    while (scanf("%d", &t)!=EOF) {
        while (t--) {
            int n;
            scanf("%d", &n);
            int cnt = 0;
            for(int i = 0; i < n; i++){
                int xx;
                scanf("%d", &xx);
                if(xx%2==1) cnt++;
            }
            printf("%0.f\n", (double)(dp[cnt][n-cnt]*3));
        }
    }
}