688A: Opponents

Codeforces Round #360 Editorial [+ Challenges!]

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

Input

4 1
0100

Output

Input

Output

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 1, 3 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.

Solution

Let's find out for each row of the given matrix if it is completely consisting of ones or not. Make another array canWin, and set canWin_i equal to one if the i-th row consists at least one zero. Then the problem is to find the maximum subsegment of canWin array, consisting only ones. It can be solved by finding for each element of canWin, the closest zero to it from left. The complexity of this solution is O(nd), but the limits allow you to solve the problem in O(nd²) by iterating over all possible subsegments and check if each one of them is full of ones or not.

C++ code

//     . .. ... .... ..... be name khoda ..... .... ... .. .     \

#include <bits/stdc++.h>
using namespace std;

inline int in() { int x; scanf("%d", &x); return x; }
const int N = 202;

int a[N][N];

int main()
{
	int n = in(), d = in();
	int ans = 0, cur = 0;
	for(int i = 0; i < d; i++)
	{
		string s;
		cin >> s;
		bool iff = 0;
		for(int j = 0; j < n; j++)
			iff |= (s[j] == '0');
		if(iff)
			cur++;
		else
			cur = 0;
		ans = max(ans, cur);
	}
	cout << ans << endl;
}

Python code

n, d = map(int, raw_input().split())
cur = 0
ans = 0
for i in range(d):
    s = raw_input()
    if (s == '1' * n):
        cur = 0
    else:
        cur += 1
    ans = max(ans, cur)
print ans

原文链接：Codeforces Round #360 Editorial [+ Challenges!] - Codeforces