C.Going Home(思维+抽屉原理)

https://codeforces.com/contest/1501/problem/C

---

思路:

当时做的时候看了值域和n,只知道会有冲撞,不会那么大,但具体构造不太会。

给出这个值域是2.5e6,开个5e6的数组,n²暴力枚举就往桶里扔。也就是cnt[a[i] + a[j]] ++;

如果这个cnt >= 2那么就找到了答案。

最差情况下循环5e6 + 1次数组中必然会出现某个位置cnt[i] >= 2

#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<cstdio>
#include<algorithm>
#define debug(a) cout<<#a<<"="<<a<<endl;
using namespace std;
const int maxn=5e6+100;
typedef long long LL;
inline LL read(){LL x=0,f=1;char ch=getchar();	while (!isdigit(ch)){if (ch=='-') f=-1;ch=getchar();}while (isdigit(ch)){x=x*10+ch-48;ch=getchar();}
return x*f;}
LL a[200010],b[maxn],c[maxn],cnt[maxn];
int main(void)
{
  cin.tie(0);std::ios::sync_with_stdio(false);
  LL n;cin>>n;
  for(LL i=1;i<=n;i++) cin>>a[i];
  for(LL i=1;i<=n;i++){
    for(LL j=i+1;j<=n;j++){
        cnt[a[i]+a[j]]++;
        if(!b[a[i]+a[j]] ) b[a[i]+a[j]]=i;
        if(!c[a[i]+a[j]] ) c[a[i]+a[j]]=j;
        if(cnt[a[i]+a[j]]>=2&&b[a[i]+a[j]]!=i&&b[a[i]+a[j]]!=j&&c[a[i]+a[j]]!=i&&c[a[i]+a[j]]!=j){
            cout<<"YES"<<"\n";
            cout<<b[a[i]+a[j]]<<" "<<c[a[i]+a[j]]<<" "<<i<<" "<<j<<"\n";
            return 0;
        }
    }
  }
  cout<<"NO"<<"\n";
return 0;
}