zoukankan      html  css  js  c++  java
  • Codeforces Round #360 (Div. 2) A. Opponents 水题

    A. Opponents

    题目连接:

    http://www.codeforces.com/contest/688/problem/A

    Description

    Arya has n opponents in the school. Each day he will fight with all opponents who are present this day. His opponents have some fighting plan that guarantees they will win, but implementing this plan requires presence of them all. That means if one day at least one of Arya's opponents is absent at the school, then Arya will beat all present opponents. Otherwise, if all opponents are present, then they will beat Arya.

    For each opponent Arya knows his schedule — whether or not he is going to present on each particular day. Tell him the maximum number of consecutive days that he will beat all present opponents.

    Note, that if some day there are no opponents present, Arya still considers he beats all the present opponents.

    Input

    The first line of the input contains two integers n and d (1 ≤ n, d ≤ 100) — the number of opponents and the number of days, respectively.

    The i-th of the following d lines contains a string of length n consisting of characters '0' and '1'. The j-th character of this string is '0' if the j-th opponent is going to be absent on the i-th day.

    Output

    Print the only integer — the maximum number of consecutive days that Arya will beat all present opponents.

    Sample Input

    2 2
    10
    00

    Sample Output

    2

    Hint

    题意

    有一个人,要和n个人pk,要pk d天

    如果这一天所有人都来了,他就输了

    否则这个人就会说胜利

    问这个人最多能够连续胜利多少天

    题解:

    直接暴力做就好了……

    水题

    代码

    #include<bits/stdc++.h>
    using namespace std;
    
    string s;
    int main()
    {
        int n,d;
        scanf("%d%d",&n,&d);
        int ans = 0, tmp = 0;
        for(int i=0;i<d;i++){
            cin>>s;
            int flag = 0;
            for(int j=0;j<s.size();j++){
                if(s[j]=='0')
                    flag = 1;
            }
            if(flag==1)tmp=tmp+1;
            else tmp=0;
            ans = max(ans,tmp);
        }
        cout<<ans<<endl;
    }
  • 相关阅读:
    Python3 list基本操作
    Oracle Sql关于case-when,if-then,decode
    Oracle12c解锁scott测试用户
    Java得到下一天日期
    SQLiteTest源代码
    实况照片的视频合并
    一个支持中文的日志类
    为对话框添加背景图片
    获取当前应用程序的文件名
    Win7系统x64正在准备再循环
  • 原文地址:https://www.cnblogs.com/qscqesze/p/5631045.html
Copyright © 2011-2022 走看看