CF1076C Meme Problem 数学-白红宇

CF1076C Meme Problem 数学

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发布时间：2019-06-25

本文共 2607 字，大约阅读时间需要 8 分钟。

Try guessing the statement from this picture:

$\text{[math]}$

You are given a non-negative integer

. You have to find two non-negative real numbers

Input

The first line contains

) — the number of test cases.

Each test case contains one integer

Output

For each test print one line.

If there is an answer for the

If there is no answer for the

-th test, print "N".

Your answer will be considered correct if

Example

Input

Copy

76901459991000

Output

Copy

Y 67.985071301 1.014928699Y 0.000000000 0.000000000NY 2.000000000 2.000000000Y 3.618033989 1.381966011Y 997.998996990 1.001003010Y 998.998997995 1.001002005 题意：求两个实数 a,b 满足 a+b==d&&a*b==d；

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                         //#include
                         
                          //#pragma GCC optimize("O3")using namespace std;#define maxn 400005#define inf 0x3f3f3f3f#define INF 9999999999#define rdint(x) scanf("%d",&x)#define rdllt(x) scanf("%lld",&x)#define rdult(x) scanf("%lu",&x)#define rdlf(x) scanf("%lf",&x)#define rdstr(x) scanf("%s",x)typedef long long ll;typedef unsigned long long ull;typedef unsigned int U;#define ms(x) memset((x),0,sizeof(x))const long long int mod = 1e9 + 7;#define Mod 1000000000#define sq(x) (x)*(x)#define eps 1e-3typedef pair
                          
                            pii;#define pi acos(-1.0)const int N = 1005;#define REP(i,n) for(int i=0;i<(n);i++)typedef pair
                           
                             pii;inline ll rd() { ll x = 0; char c = getchar(); bool f = false; while (!isdigit(c)) { if (c == '-') f = true; c = getchar(); } while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); } return f ? -x : x;}ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a%b);}ll sqr(ll x) { return x * x; }/*ll ans;ll exgcd(ll a, ll b, ll &x, ll &y) { if (!b) { x = 1; y = 0; return a; } ans = exgcd(b, a%b, x, y); ll t = x; x = y; y = t - a / b * y; return ans;}*/ll qpow(ll a, ll b, ll c) { ll ans = 1; a = a % c; while (b) { if (b % 2)ans = ans * a%c; b /= 2; a = a * a%c; } return ans;}int T;int d;int main(){ //ios::sync_with_stdio(0); rdint(T); while (T--) { rdint(d); if (d*d < 4 * d)cout << "N" << endl; else { double a = 1.0*(d + sqrt(d*d - 4 * d)) / (2.0); double b = 1.0*d - a; cout << "Y" << ' '; printf("%.9lf %.9lf\n", 1.0*a, 1.0*b); } } return 0;}