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  • codeforces 688A A. Opponents(水题)

    题目链接:

    A. Opponents

    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    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.

    Examples
    input
    2 2
    10
    00
    output
    2
    input
    4 1
    0100
    output
    1
    input
    4 5
    1101
    1111
    0110
    1011
    1111
    output
    2

    题意:

    全是1的时候就是失败的时候,找最长的0的序列;

     思路:

     水题

     AC代码:

    //#include <bits/stdc++.h>
    #include <vector>
    #include <iostream>
    #include <queue>
    #include <cmath>
    #include <map>
    #include <cstring>
    #include <algorithm>
    #include <cstdio>
    
    using namespace std;
    #define Riep(n) for(int i=1;i<=n;i++)
    #define Riop(n) for(int i=0;i<n;i++)
    #define Rjep(n) for(int j=1;j<=n;j++)
    #define Rjop(n) for(int j=0;j<n;j++)
    #define mst(ss,b) memset(ss,b,sizeof(ss));
    typedef  long long LL;
    template<class T> void read(T&num) {
        char CH; bool F=false;
        for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
        for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
        F && (num=-num);
    }
    int stk[70], tp;
    template<class T> inline void print(T p) {
        if(!p) { puts("0"); return; }
        while(p) stk[++ tp] = p%10, p/=10;
        while(tp) putchar(stk[tp--] + '0');
        putchar('
    ');
    }
    
    const LL mod=1e9+7;
    const double PI=acos(-1.0);
    const LL inf=1e18;
    const int N=1e4+20;
    const int maxn=1005;
    const double eps=1e-10;
    
    char str[103][103];
    int a[103];
    int main()
    {
    
        int n,d;
        read(n);read(d);
        for(int i=1;i<=d;i++)
        {
            scanf("%s",str[i]);
            int flag=0;
            for(int j=0;j<n;j++)
            {
                if(str[i][j]=='0')flag=1;
            }
            if(flag)a[i]=1;
        }
        int ans=0,len=0;
        for(int i=1;i<=d;i++)
        {
            if(a[i])
            {
                len++;
            }
            else
            {
                ans=max(ans,len);
                len=0;
            }
        }
        ans=max(ans,len);
            cout<<ans<<"
    ";
    
            return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zhangchengc919/p/5629313.html
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