CF688A Opponents_代码007(未授权)

本文介绍: 小白有 n 个对手，他每天都要和这些对手PK。对于每一天，如果 n 个对手全部到齐，那么小白就输了一场，否则小白就赢了一场。特别的，如果某天一个对手都没有到，也算小白赢。现在已知对手 d 天的出场情况，请计算小白最多能连胜多少场。

小白有 n 个对手，他每天都要和这些对手PK。对于每一天，如果 n 个对手全部到齐，那么小白就输了一场，否则小白就赢了一场。特别的，如果某天一个对手都没有到，也算小白赢。现在已知对手 d 天的出场情况，请计算小白最多能连胜多少场。

第一行，两个整数 n , d ( 1 ≤ n,d ≤ 100 )。接下来 d 行，每行 n 个 0 或 1 的整数，依次表示这一天所有对手的到场情况， 1 表示到场， 0 表示缺席。

一个整数，表示最多的连胜场次。

Ary a ha s $ n $ opponents in th e school. Ea ch day he will fi g ht wi th all opponents wh o a re present th i s day. Hi s opponents ha v e some fi g h ting plan t hat guar antee s th e y will win, bu t implem enting th is plan requires prese nc e of th em all. That m eans if one day at le ast one of Arya’s opponents is absent at the school, then Arya will beat all present opponents. Othe r wise, if all opponents are p resent, then the y will beat Ar ya.

For each opponent Arya knows his schedule — wh ether or not he is go in g to p resent on each particular day. Tell him the maximum number of consecutive days that he will beat all p resent opponents.

Note, that if some day there are no opponents p resent, Arya still conside rs he beats all the p resent opponents.

2 2
10
00

4 1
0100

//
// Created by Gowi on 2023/12/2.
//

#include <iostream>
#include <string>
#include <algorithm>

#define N 120

using namespace std;

int main() {
    int n, d;
    string opponents[N];
    cin >> n >> d;
    for (int i = 0; i < d; ++i) {
        cin >> opponents[i];
    }
    int t = 0, res = 0;
    for (int i = 0; i < d; ++i) {
        int flag = 1;
        for (int j = 0; j < n; ++j) {
            if (opponents[i][j] == '0') {
                flag = 0;
                t++;
                break;
            }
        }
        if (flag == 1) {
            t = 0;
        }
        res = max(res, t);
    }
    cout << res << endl;
    return 0;
}