CF1076C Meme Problem 数学-CFANZ编程社区

Try guessing the statement from this picture:

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≤103

) — the number of test cases.

Each test case contains one integer d

(0≤d≤103)

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

and |(a+b)−d|≤10−6

Example Input Copy

Output Copy

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

题意：求两个实数 a,b 满足 a+b==d&&a*b==d；

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
//#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-3
typedef 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;
}

EPFL - Fighting