IndiaHacks 2016 - Online Edition (Div. 1 + Div. 2)-A - Bear and Three Balls-水题

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http://codeforces.com/contest/653/problem/A

直接判断有没连在一起的3个数就好了

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <map>#include <set>#include <vector>#include <iostream>using namespace std;const double pi=acos(-1.0);double eps=0.000001; __int64 vis1[1000006];__int64 vis2[1000006];  __int64 when1[1000006];__int64 when2[1000006]; __int64 gcdc(__int64 a,__int64 b){      if(b==0)          return a;   return gcdc(b,a%b);  }  __int64 t,p,c;void exgcd(__int64 a,__int64 b,__int64&d,__int64&x,__int64&y)//ax+by=gcd(a,b){    if(!b)    {        d=a;        x=1;        y=0;    }    else    {        exgcd(b,a%b,d,y,x);        y-=x*(a/b);    }}int vis[1888];int main(){  int i,x; int n; cin>>n; for (i=1;i<=n;i++) {cin>>x;vis[x]=1; } int flag=0; for (i=1;i<=1000;i++) { if(vis[i]&&vis[i+1]&&vis[i+2]) {flag=1;break;} } if (flag) printf("YES\n"); else printf("NO\n");return 0;}

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