Opponents

題意：一共有d天，每天和n個人打架，如果某天n個人有人沒有出現，那麽可以打贏，問最多連續打贏幾天，輸入d個字符串，第i位為0表示第i個人沒有出現，為1表示出現了

思路：xjb寫

AC代碼：

#include "iostream"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h
 
"
#pragma comment(linker, "/STACK:102400000,102400000")
#define ll long long
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a,x) memset(a,x,sizeof(a))
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define lrt (rt<<1)
#define rrt (rt<<1|1)
using namespace std;
 
const long long INF = 1e18+1LL;
const int inf = 1e9+1e8;
const int N=1e5+100;
const ll mod=1e9+7;

///AAAA
char s[105];
int d,n;
map<int,int> M;
int main(){
    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>n>>d;
    int l;
    int ans=0,m=0;
    for(int k=1; k<=d; ++k){
        
 
int f=0;
        cin>>s+1;
        for(int i=1; i<=n; ++i){
            if(s[i]==‘0‘){
                f=1;
                break;
            }
        }
        if(f){
            m++;
            ans=max(ans,m);
        }
        else m=0;
    }
    cout<<ans<<"\n";
    return 0;

}

Codeforces Round #360 A