您的位置：首页 > 其它

CodeForces 688A

2016-07-05 23:46 344 查看

http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=407720

Time Limit: 1000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

Submit StatusDescriptionArya 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 implementingthis 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.InputThe 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 consistingof characters '0' and '1'. The j-th character of this string is '0' if the j-thopponent is going to be absent on the i-th day.OutputPrint the only integer — the maximum number of consecutive days that Arya will beat all present opponents.Sample InputInput

2 2
10
00

Output

Input

4 1
0100

Output

Input

4 511011111011010111111

Output

SourceCodeforces Round #360 (Div. 2)

#include<iostream>
#include<iomanip>
#include<algorithm>
#include<istream>
#include<ostream>
#include<cstring>
#include<cstdio>
using namespace std;
typedef long long ll;
char a[110][110];
int b[110];
int main(){
int n,d;
while(~scanf("%d%d",&n,&d)){
memset(b,0,sizeof(b));
for(int i=0;i<d;i++){
scanf("%s",&a[i]);
}
for(int i=0;i<d;i++){
for(int k=0;k<n;k++){
if(a[i][k]==48){
b[i]=1;
break;
}
}
}

int maxn=0;//@1for(int i=0;i<d;++i){int cnt=0;for(int j=i;j<d;j++){if(b[j]){cnt++;}else break;}maxn=max(maxn,cnt);}cout<<maxn<<endl;}return 0;}

@1:

内容来自用户分享和网络整理，不保证内容的准确性，如有侵权内容，可联系管理员处理

标签：

相关文章推荐

新的分享

章节导航