2011 Multi-University Training Contest 1 - Host by HNU&&YY's new problem
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刚开始看这道题时便发现时间控制为4秒,如果按一般的方法三层for循环,,果断tle,因此想一定用到一点高级算法,,然后想到了hash表,把每次出现的数标记为1,一开始把全部数标记为0,对每次输入的数进行对称查找hash表,假设输入的数为a,如果出现hash[a-i]+hash[a+i]=1则说明一定存在,,,
AC代码:
#include<iostream>#include<string.h>using namespace std;int hash[10001];int main(){ int n; cin>>n; while(n--) { int m; cin>>m;memset(hash,0,sizeof(hash));int a,i,j;bool flag=false;for(i=0;i!=m;++i){ cin>>a; hash[a]++; if(!flag) { for(j=1;j<a&&a+j<=m;++j) if(hash[a-j]+hash[a+j]==1) {flag=true;break;} }}if(flag) cout<<"Y"<<endl;else cout<<"N"<<endl; }return 0;}
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