cf Educational Codeforces Round 54 C. Meme Problem

原题：
C. Meme Problem
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output

You are given a non-negative integer d. You have to find two non-negative real numbers a and b such that a+b=d and a⋅b=d.

Input
The first line contains t (1≤t≤10^3) — the number of test cases.

Each test case contains one integer d (0≤d≤10^3).

Output
For each test print one line.

If there is an answer for the i-th test, print “Y”, and then the numbers a and b.

If there is no answer for the i-th test, print “N”.

Your answer will be considered correct if ∣(a+b)−a⋅b∣≤10−6|(a+b)−a⋅b|≤10^{−6}∣(a+b)−a⋅b∣≤10−6 and ∣(a+b)−d∣≤10−6|(a+b)−d|≤10^{−6}∣(a+b)−d∣≤10−6.

Example
inputCopy
7
69
0
1
4
5
999
1000
output
Y 67.985071301 1.014928699
Y 0.000000000 0.000000000
N
Y 2.000000000 2.000000000
Y 3.618033989 1.381966011
Y 997.998996990 1.001003010
Y 998.998997995 1.001002005

中文：

给你一个数d，让你找出两个数a和b，使得a+b=d,a×b=d，如果存在这两个数，输出Y然后输出a和b，否则输出N

代码：

#include <bits/stdc++.h>using namespace std;
typedef long long ll;
typedef pair<int,int> pii;int t;
double d,a,b;
int main()
{ios::sync_with_stdio(false);cin>>t;while(t--){cin>>d;if(d*d-4*d<0){cout<<"N"<<endl;continue;}double tmp=sqrt(d*d-4*d);cout<<fixed<<setprecision(9)<<"Y"<<" "<<(d+tmp)/2<<" "<<(d-tmp)/2<<endl;}return 0;
}

解答：

这题也太简单了吧-_-|||
2元一次方程求解的问题，上过初中的应该都能解决

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