Codeforces-688A Opponents

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.

题目大意:

题目说的就是一个人,有n个对手,当n个对手都出席的时候,这个人被打败,一旦有人没出席,这个人就会打败所有人,问你d天他有多少天是连续打败所有人的。

解法:

比较水= =就直接上代码了。

#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 100 + 5;
int main()
{
	int n, d, ans = 0, tmp = 0;
	scanf("%d%d", &n, &d);
	for(int i = 0; i < d; ++i){
		char str[maxn] = {0};
		scanf(" %s", str);
		bool flag = false;
		for(int j = 0; j < n; ++j){
			if(str[j] == '0') {
				flag = true;
				break;
			}
		}
		if(flag){
			tmp ++;
		}else{
			tmp = 0;
		}
		if(tmp > ans) ans = tmp;
	}
	printf("%d\n", ans);
	return 0;
}


### Codeforces Problem 976C Solution in Python For solving problem 976C on Codeforces using Python, efficiency becomes a critical factor due to strict time limits aimed at distinguishing between efficient and less efficient solutions[^1]. Given these constraints, it is advisable to focus on optimizing algorithms and choosing appropriate data structures. The provided code snippet offers insight into handling string manipulation problems efficiently by customizing comparison logic for sorting elements based on specific criteria[^2]. However, for addressing problem 976C specifically, which involves determining the winner ('A' or 'B') based on frequency counts within given inputs, one can adapt similar principles of optimization but tailored towards counting occurrences directly as shown below: ```python from collections import Counter def determine_winner(): for _ in range(int(input())): count_map = Counter(input().strip()) result = "A" if count_map['A'] > count_map['B'] else "B" print(result) determine_winner() ``` This approach leverages `Counter` from Python’s built-in `collections` module to quickly tally up instances of 'A' versus 'B'. By iterating over multiple test cases through a loop defined by user input, this method ensures that comparisons are made accurately while maintaining performance standards required under tight computational resources[^3]. To further enhance execution speed when working with Python, consider submitting codes via platforms like PyPy instead of traditional interpreters whenever possible since they offer better runtime efficiencies especially important during competitive programming contests where milliseconds matter significantly.
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