• D. Ticket Game


    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.

    Input

    The first line contains one even integer n(2n2105)(2≤n≤2⋅105) — the number of digits in the ticket.

    The second line contains a string of nn 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.

    Output

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

    Examples
    input
    Copy
    4
    0523
    
    output
    Copy
    Bicarp
    
    input
    Copy
    2
    ??
    
    output
    Copy
    Bicarp
    
    input
    Copy
    8
    ?054??0?
    
    output
    Copy
    Bicarp
    
    input
    Copy
    6
    ???00?
    
    output
    Copy
    Monocarp
    
    Note

    Since there is no question mark in the ticket in the first example, the winner is determined before the game even starts, and it is Bicarp.

    In the second example, Bicarp also wins. After Monocarp chooses an erased digit and replaces it with a new one, Bicap can choose another position with an erased digit and replace it with the same digit, so the ticket is happy.

    关键在于一个等号

    #include <iostream>
    #include <vector>
    #include <algorithm>
    #include <string>
    #include <set>
    #include <queue>
    #include <map>
    #include <sstream>
    #include <cstdio>
    #include <cstring>
    #include <numeric>
    #include <cmath>
    #include <iomanip>
    #include <deque>
    #include <bitset>
    #include <unordered_set>
    #include <unordered_map>
    #define ll              long long
    #define PII             pair<int, int>
    #define rep(i,a,b)      for(int  i=a;i<=b;i++)
    #define dec(i,a,b)      for(int  i=a;i>=b;i--)
    using namespace std;
    int dir[4][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 } };
    const long long INF = 0x7f7f7f7f7f7f7f7f;
    const int inf = 0x3f3f3f3f;
    const double pi = 3.14159265358979323846;
    const double eps = 1e-6;
    const int mod =1e9+7;
    const int N = 100005;
    //if(x<0 || x>=r || y<0 || y>=c)
    
    inline ll read()
    {
        ll x = 0; bool f = true; char c = getchar();
        while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
        while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
        return f ? x : -x;
    }
    ll gcd(ll m, ll n)
    {
        return n == 0 ? m : gcd(n, m % n);
    }
    ll lcm(ll m, ll n)
    {
        return m * n / gcd(m, n);
    }
    ll qpow(ll m, ll k, ll mod)
    {
        ll res = 1, t = m;
        while (k)
        {
            if (k & 1)
                res = res * t % mod;
            t = t * t % mod;
            k >>= 1;
        }
        return res;
    }       
    
    int main()
    {
        int n;
        cin >> n;
        string s;
        cin >> s;
        int prew = 0, sufw = 0,pre=0,suf=0;
        for (int i = 0; i < n; i++)
        {
            if (i < n / 2)
            {
                if (s[i] == '?')
                    prew++;
                else
                    pre+= s[i] - '0';
            }
            else
            {
                if (s[i] == '?')
                    sufw++;
                else
                    suf+=s[i]-'0';
            }
        }
        if (pre < suf)
            swap(pre, suf), swap(prew, sufw);
        if (pre != suf)//这个位置是关键
        {
            pre += 9 * min(prew, (prew + sufw) / 2);
            suf += 9 * min(sufw, (prew + sufw) / 2);
            if (pre == suf)
                cout << "Bicarp" << endl;
            else
                cout << "Monocarp" << endl;
        }
        else
        {
            if(prew!=sufw)
                cout << "Monocarp" << endl;
            else
                cout << "Bicarp" << endl;
        }
        return 0;
    }
  • 相关阅读:
    【转载】爸妈,求你们不要这么懂事
    【Selenium】各种方式在选择的时候应该怎么选择
    java动态绑定与静态绑定【转】
    java 字符串池【转】
    【转】java中PriorityQueue优先级队列使用方法
    【转】程序员的出路之一
    关于现在,关于未来
    计算机网络实验(router_sim)工具
    MIT算法导论课程
    最大间隙问题
  • 原文地址:https://www.cnblogs.com/dealer/p/13088950.html
Copyright © 2020-2023  润新知