标签：inpu fir 奇数 intended special sed return ber lse
Poor Rukaw
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 223 Accepted Submission(s):
102
Problem Description
Last time, Hanamichi lost the basketball battle between
him and Rukaw. So these days he had a burning desire to wreak revenge. So he
invented a new game and asked Rukaw to play with him. What’s more, the loser
should confess his ignorance and on the other hand the winner can go on a trip
with Haruko.
Hanamichi knows the game well (as he invented it), and he
always chooses the best strategy, however Rukaw is not so willing to play this
game. He just wants it to be finished as soon as possible so that he can go to
play basketball. So he decides not to think about the best strategy and play
casually.
The game’s rules are here. At first, there are N numbers on the
table. And the game consists of N rounds. Each round has a score which is the
number of the numbers still left on the table. And Each round there will be one
number to be removed from the table. In each round, two players take turns to
play with these numbers.To be fair, Rukaw plays first in the first round. If
there’s more than 1 numbers on the table, players can choose any two numbers
they like and change them to a number abs(x-y). This round ends when there’s
only one number left on the table, and if this number is an odd number, Rukaw
wins, otherwise Hanamichi wins. The score of this round will be add to the
winner. After that, all numbers will be recovered to the state when this round
starts. And the loser of this round has the right to remove one number and he
also has the right to play first in the next round. Then they use the remaining
numbers to start next round. After N rounds, all numbers removed and this game
ends. The person who has more scores wins the whole game.
As you know,
Rukaw has already decided to play casually, that is to say, in his turn, he
chooses numbers randomly, each numbers left on the table has the same
possibility to be chosen. When a round ends, if Rukaw is the loser, he also
randomly chooses a number to remove. And Hanamichi will always choose numbers or
remove numbers to maxmium his final total score. Here comes the
question:
Given the N numbers on the table at the beginning, can you
calculate the expectation of the final score of Hanamichi. (We don’t care about
who wins the whole game at all.)
Input
This problem contains multiple tests.
In the first
line there’s one number T (1 ≤ T ≤ 200) which tells the total number of test
cases. For each test case, there a integer N (1 ≤ N ≤ 1000) in the first line,
and there are N intergers Ai , i = 1, 2, … , N (1 ≤ Ai ≤ 100000), in the second
line which are the numbers at the beginning.
Output
This problem is intended to use special judge. But so
far, BestCoder doesn’t support special judge. So you should output your answer
in the following way. If the expectation you get is X, output [3×X+0.5]![技术图片]()
in a line. Here, [A] means the largest integer which is no more than A.
Sample Input
Sample Output
9
3
析：桌子上有n个数，两个玩家甲和乙进行n轮游戏，每轮游戏的分数为该轮数字个数，甲随便玩，乙采取最优策略，当桌面数字多于2个时，玩家可任选两个数进行|(X-Y)|运算，并将结果放入桌面剩下的数中，一直到只剩下一个数，为奇数，甲赢，否则乙赢，输的选手可在下一轮开始时从桌面任意移除一个数字并获得优先进行游戏的权力，问最终乙的得分期望
对甲来说，最后一轮为奇数时是必胜的，对于每一轮的转移，有 拿走两个奇数变为偶数，一奇一偶变奇数，两偶数变偶数，可知奇数的变化只能为0或2，所以如果开始有奇数个奇数，那么甲一定获胜
因此，判断每一轮中奇数个数，若当前轮奇数个数为奇数，那么乙输，一定会拿走一个奇数，改变当前状态，否则甲输，随机选取，即求选取奇数和偶数的概率
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#define ll long long
using namespace std;
const int N = 1005;
int t, n, x;
double dp[N][N];
double dfs(int o, int e){
if(o+e <= 0)
return 0;
if(dp[o][e]) //剪枝
return dp[o][e];
if(o&1) // 奇数个数为奇数，Hanamichi输，拿走一个奇数
return dp[o][e] = dfs(o-1, e);
if(o){
if(e) // 奇数个数为偶数，Rukaw输，随机拿走一个
return dp[o][e] = (o*dfs(o-1, e)+e*dfs(o, e-1))/(o+e)+o+e;
else
return dp[o][e] = dfs(o-1, e)+o;
}
return dp[o][e] = dfs(o, e-1)+e;
}
int main(){
scanf("%d", &t);
while(t--){
scanf("%d", &n);
int odd = 0, even = 0;
memset(dp, 0, sizeof(dp));
for(int i = 0; i < n; i ++){
scanf("%d", &x);
if(x&1)
odd ++;
else
even ++;
}
int res = 3*dfs(odd, even)+0.5;
printf("%d\n",res);
}
return 0;
}
View Code
hdu 4958（博弈论+概率DP）
标签：inpu fir 奇数 intended special sed return ber lse
原文地址：https://www.cnblogs.com/microcodes/p/12876961.html
