灰魔法师（暴力）

思路：枚举可能的完全平方数，用个map存储a[i]的出现次数，看每个数对完全平方数的贡献次数

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include<iostream>
#include<vector>
#include<queue>
//#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define SIS std::ios::sync_with_stdio(false)
#define space putchar(' ')
#define enter putchar('\n')
#define lson root<<1
#define rson root<<1|1
typedef pair<int,int> PII;
const int mod=998244353;
const int N=2e6+10;
const int M=2e3+10;
const int inf=0x7f7f7f7f;
const int maxx=2e5+7;

ll gcd(ll a,ll b)
{
    return b==0?a:gcd(b,a%b);
}

ll lcm(ll a,ll b)
{
    return a*(b/gcd(a,b));
}

template <class T>
void read(T &x)
{
    char c;
    bool op = 0;
    while(c = getchar(), c < '0' || c > '9')
        if(c == '-')
            op = 1;
    x = c - '0';
    while(c = getchar(), c >= '0' && c <= '9')
        x = x * 10 + c - '0';
    if(op)
        x = -x;
}
template <class T>
void write(T x)
{
    if(x < 0)
        x = -x, putchar('-');
    if(x >= 10)
        write(x / 10);
    putchar('0' + x % 10);
}
ll qsm(int a,int b,int p)
{
    ll res=1%p;
    while(b)
    {
        if(b&1) res=res*a%p;
        a=1ll*a*a%p;
        b>>=1;
    }
    return res;
}

ll mp[N];
int a[N];

int main()
{
   int n;
   scanf("%d",&n);
   for(int i=1;i<=n;i++)
   {
      // int x;
       scanf("%d",&a[i]);
       mp[a[i]]++;
   }
   sort(a+1,a+1+n);
   ll ans=0;
   for(int i=2;i*i<=200000;i++)
   {
       for(int j=1;j<=n;j++)
       {
           int num=i*i-a[j];
           if(num<a[j])continue;
           if(a[j]==num) ans+=mp[a[j]]*(mp[a[j]]-1)/2;
           else if(mp[num]) ans+=mp[a[j]]*mp[num];
           while(a[j]==a[j+1])j++;
       }
   }
   printf("%lld",ans);

    return 0;
}