题意:一共有d天,每天和n个人打架,如果某天n个人有人没有出现,那么可以打赢,问最多连续打赢几天,输入d个字符串,第i位为0表示第i个人没有出现,为1表示出现了
思路:xjb写
AC代码:
#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; }
时间: 2024-10-29 19:13:42