思路讲解
用了一个map来存一下每个数出现的次数，然后只有出现两次的数才可参与矩形的构建。
然后C++的map是有序的，所以遍历也是有序的
AC代码
https://codeforces.com/contest/2038/submission/293462817
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| #include <iostream>
#include <algorithm>
#include <map>
#include <vector>
using namespace std;
typedef long long ll;
ll T;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
cin>>T;
while (T--) {
ll n;
cin>>n;
vector<ll> A(n+10);
map<ll,ll> occur;
for(int i=1;i<=n;++i){
cin>>A[i];
if(occur.count(A[i]))
occur[A[i]]+=1;
else
occur[A[i]]=1;
}
vector<ll> validNum;
for(map<ll,ll>::iterator it=occur.begin();it!=occur.end();it++){
ll num=it->first,cnt=it->second;
while (cnt>=2){
validNum.push_back(num);
cnt-=2;
}
}
if(validNum.size()<4){
cout<<"NO"<<endl;
}else {
cout<<"YES"<<endl;
ll a=validNum[0],b=validNum[1],c=validNum[validNum.size()-1],d=validNum[validNum.size()-2];
cout<<a<<" "<<b<<" "<<a<<" "<<c<<" "<<d<<" "<<b<<" "<<d<<" "<<c<<endl;
}
}
}
|
心路历程（WA，TLE，MLE……）
输出顺序要好好想想
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| cout<<a<<" "<<b<<" "<<a<<" "<<c<<" "<<d<<" "<<b<<" "<<d<<" "<<c<<endl;
|