1857F. Sum and Product（转化题目条件） · znzryb
## 思路讲解
题意还是比较简单的，告诉你两数之和以及两数之积，问你在A数组中有几个对满足这个条件。
感觉是数据结构，但没想到要对题目条件进行转化。
看了一眼题解，好像是要转化题目条件。有点像下面这道题目，但没有那么明显。
[2072E. Do You Love Your Hero and His Two-Hit Multi-Target Attacks?](/2025/03/08/20250308-e9d462b1)
又看了一眼，竟然是韦达定理？
离谱，我是完全没想到啊，这个我都已经不太记得了。
韦达定理+求根公式
## AC代码
[https://codeforces.com/problemset/submission/1857/310498477](https://codeforces.com/problemset/submission/1857/310498477)
```cpp
// Problem: F. Sum and Product
// Contest: Codeforces - Codeforces Round 891 (Div. 3)
// URL: https://codeforces.com/problemset/problem/1857/F
// Memory Limit: 256 MB
// Time Limit: 4000 ms
// by znzryb
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define FOR(i, a, b) for (long long i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (long long i = (a); i >= (b); --i)
#define all(x) x.begin(),x.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> pll;
typedef array<ll,4> arr;
typedef double DB;
typedef pair<DB,DB> pdd;
constexpr ll MAXN=static_cast<ll>(2e5)+10,INF=static_cast<ll>(5e18)+3;
constexpr double eps=1e-10;
ll N,T,A[MAXN];
inline void solve(){
cin>>N;
for(int i=1;i<=N;++i){
cin>>A[i];
}
sort(A+1,A+1+N);
ll Q;
cin>>Q;
for(int i=1;i<=Q;++i){
ll b,c;
cin>>b>>c;
b=-b;
ll D=b*b-4*1*c;
if(D<0){
cout<<0<<" ";
continue;
}
long double sqD=sqrt(D);
// #ifdef LOCAL
// cerr<<D<<"\n";
// cerr <<setprecision(15)<< sqD << '\n';
// #endif
// if(fabs(sqD-floor(sqD))>eps){ // 判断D是不是平方数
// cout<<0<<" ";
// continue;
// }
ll fsqD=floor(sqD);
if(fsqD*fsqD!=D){
cout<<0<<" ";
continue;
}
if(D!=0){
if((-b+fsqD)%2!=0 || (-b-fsqD)%2!=0){
cout<<0<<" ";
continue;
}
ll x1=(-b+fsqD)/(2);
ll x2=(-b-fsqD)/(2);
ll num1=upper_bound(A+1,A+1+N,x1)-lower_bound(A+1,A+1+N,x1);
ll num2=upper_bound(A+1,A+1+N,x2)-lower_bound(A+1,A+1+N,x2);
cout<<num1*num2<<" ";
// #ifdef LOCAL
// cerr << num1<<" "<<num2<<" "<<x1<<" "<<x2 << '\n';
// #endif
}else{
ll needFind=(-b)/(2);
ll num=upper_bound(A+1,A+1+N,needFind)-lower_bound(A+1,A+1+N,needFind);
cout<<num*(num-1)/2<<" ";
}
}
cout<<"\n";
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
cin>>T;
while(T--){
solve();
}
return 0;
}
/*
AC
https://codeforces.com/problemset/submission/1857/310498477
*/
```
## 心路历程（WA，TLE，MLE……）
WA
[https://codeforces.com/problemset/submission/1857/310497474](https://codeforces.com/problemset/submission/1857/310497474)
被卡了精度，无视精度，直接用整数比较好。
```cpp
ll fsqD=floor(sqD);
if(fsqD*fsqD!=D){
cout<<0<<" ";
continue;
}
```