Codeforces Round #360 (Div. 2) A. Opponents

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
Note

In the first and the second samples, Arya will beat all present opponents each of the d days.

In the third sample, Arya will beat his opponents on days 13 and 4 and his opponents will beat him on days 2 and 5. Thus, the maximum number of consecutive winning days is 2, which happens on days 3 and 4.




题意
n个人,m天,接下来接收每天每个人是否在学校,1表示在学校,0表示不在,如果所有人都在学校那么Arya打不过他们,如果人不全或者没有人那么认为Arya能打赢,让你输出Arya最大的连胜天数。

代码
#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n,i,a,b,m,x,flag;
    char z[105][105];
    m=x=0;
    cin>>n>>i;
    for(a=0;i>a;a++){
        cin>>z[a];
    }
    for(a=0;i>a;a++){
        flag=0;
        for(b=0;n>b;b++){
            if(z[a][b]=='0'){
                x++;
                flag=1;
                break;
            }
        }
        m=max(m,x);
        if(flag==0){
            x=0;
        }
    }
    cout<<m<<endl;
    return 0;
}


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转载自blog.csdn.net/ontheway0101/article/details/51788853