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Ticket Game CodeForces - 1215D 博弈题

liuchanglc 2020-04-23 15:55 原文

题目描述

Monocarp and Bicarp live in Berland, where every bus ticket consists of n digits (n is an even number). During the evening walk Monocarp and Bicarp found a ticket where some of the digits have been erased. The number of digits that have been erased is even.

Monocarp and Bicarp have decided to play a game with this ticket. Monocarp hates happy tickets, while Bicarp collects them. A ticket is considered happy if the sum of the first \(\frac{n}{2}\) digits of this ticket is equal to the sum of the last \(\frac{n}{2}\) digits.

Monocarp and Bicarp take turns (and Monocarp performs the first of them). During each turn, the current player must replace any erased digit with any digit from 0 to 9. The game ends when there are no erased digits in the ticket.

If the ticket is happy after all erased digits are replaced with decimal digits, then Bicarp wins. Otherwise, Monocarp wins. You have to determine who will win if both players play optimally.

输入

The first line contains one even integer n (2≤n≤2⋅105) — the number of digits in the ticket.

The second line contains a string of n digits and "?" characters — the ticket which Monocarp and Bicarp have found. If the ii-th character is "?", then the ii-th digit is erased. Note that there may be leading zeroes. The number of "?" characters is even.

输出

If Monocarp wins, print "Monocarp" (without quotes). Otherwise print "Bicarp" (without quotes).

样例

输入样例

Input

4
0523

Output

Bicarp

Input

2
??

Output

Bicarp

Input

8
?054??0?

Output

Bicarp

Input

6
???00?

Output

Monocarp

分析

一句话题意：一张票有n位数，如果这张票的前一半数字的和等于后一半数字的和(n一定是偶数)，就称这张票为快乐票。有些数被擦除了，标记为’?’(’?‘的个数也是偶数)，现在Monocarp 和 Bicarp 进行一个游戏，两人轮流将’?'变换成0到9的任意一个数，Monocarp先手，如果最后票为快乐票则Bicarp赢，否则Monocarp赢。

我们分情况来考虑一下

1、左边的数字之和等于右边的数字之和

这时，如果左右两边？的个数相等的话，后手赢，因为先手无论放什么数，后手都可以放一个相同的数来平衡

如果不相等，则先手必胜，因为最后肯定只能在一边放，只要先手能放，就一定会打破平衡

2、左边的数字之和小于右边的数字之和

如果左边？的个数大于等于右边的个数，那么先手必胜，因为先手可以一直在左边放9，后手只能不断在右边放9维持差距，但始终不能弥补差距

如果左边？的个数小于右边的个数，我们设左边？的个数为a，右边？的个数为b，那么前2a回合，策略同上；2a回合之后，先手放一个数x，后手唯一的选择就是放一个y，使x+y=9，所以当左右两边数字之差为9的倍数时，后手胜，否则先手胜

3、右边的数字之和小于左边的数字之和（同上）

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
using namespace std;
char c[300000];
int lefw,rigw,lef,rigt;
int main(){
    int n;
    scanf("%d",&n);
    getchar();
    for(int i=1;i<=n/2;i++){
        scanf("%c",&c[i]);
        if(c[i]=='?'){
            lefw++;
        } else {
            lef+=(c[i]-'0');
        }
   }
   for(int i=n/2+1;i<=n;i++){
        scanf("%c",&c[i]);
        if(c[i]=='?'){
            rigw++;
        } else {
            rigt+=(c[i]-'0');
        }
   }
   if(lef==rigt){
        if(lefw==rigw) printf("Bicarp\n");
        else printf("Monocarp\n");
    }
    else if(lef>rigt){
       int cha=rigw-lefw;
       if(cha<=0){
            printf("Monocarp\n");
            return 0;
       }
       int chaa=lef-rigt;
       if(cha%2==0 && cha/2*9==chaa) printf("Bicarp\n");
        else printf("Monocarp\n");
    } else {
        int cha=lefw-rigw;
       if(cha<=0){
            printf("Monocarp\n");
            return 0;
       }
       int chaa=rigt-lef;
       if(cha%2==0 && cha/2*9==chaa) printf("Bicarp\n");
        else printf("Monocarp\n");
    }
    return 0;
}

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