CF 1215 D Ticket Game （博弈）

 链接：https://codeforces.com/problemset/problem/1215/D

Monocarp and Bicarp live in Berland, where every bus ticket consists of nn digits (nn 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 n2n2 digits of this ticket is equal to the sum of the last n2n2 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 00 to 99. 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.

Output

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

 

#include <bits/stdc++.h>
using namespace std;

const int maxn=2e5+5;
int n;
char s[maxn];
int ls, rs, x, y;

int main()
{
    scanf("%d", &n);
    scanf("%s", s+1);
    for(int i=1; i<=n/2; i++){
        if(s[i]==‘?‘) x++;
        else ls+=s[i]-‘0‘;
    }
    for(int i=n/2+1; i<=n; i++){
        if(s[i]==‘?‘) y++;
        else rs+=s[i]-‘0‘;
    }
    if(ls<rs){
        swap(ls, rs); swap(x, y);
    }
    if(ls==rs){
        printf("%s\n", x==y? "Bicarp":"Monocarp");
    }
    else{
        if(x>=y) printf("Monocarp\n");
        else printf("%s\n", (y-x)%2==0 && ls-rs==(y-x)/2*9? "Bicarp":"Monocarp");
    }
    return 0;
}

View Code

 

 

CF 1215 D Ticket Game （博弈）

原文：https://www.cnblogs.com/Yokel062/p/11615314.html