#136103 (C++14) No.456 Millions of Submits!

• 問題: No.456 Millions of Submits!
• ユーザー: paruki
• 提出日時: 2016-12-07 23:55:49
• 言語: C++14 (gcc 13.3.0 + boost 1.87.0)
• 結果: OLE (最新)
  AC
  (最初)
• 実行時間: -
• コード長: 1,991 bytes
• コンパイル時間: 1,541 ms
• コンパイル使用メモリ: 168,112 KB
• 実行使用メモリ: 6,944 KB
• 最終ジャッジ日時: 2024-06-23 04:27:31
• 合計ジャッジ時間: 7,403 ms
• ジャッジサーバーID （参考情報）: judge1 / judge3

ファイルパターン	結果
other	AC * 12 OLE * 1

ソースコード

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#include "bits/stdc++.h"
using namespace std;
#define FOR(i,j,k) for(int (i)=(j);(i)<(int)(k);++(i))
#define rep(i,j) FOR(i,0,j)
#define each(x,y) for(auto &(x):(y))
#define mp make_pair
#define mt make_tuple
#define all(x) (x).begin(),(x).end()
#define debug(x) cout<<#x<<": "<<(x)<<endl
#define smax(x,y) (x)=max((x),(y))
#define smin(x,y) (x)=min((x),(y))
#define MEM(x,y) memset((x),(y),sizeof (x))
#define sz(x) (int)(x).size()
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<ll> vll;
/**
b = 0のとき
n^a = t
a!=0なので
より
n = t^(1/a)
(a,b) = (0,1)
log(n) = t >= 1.0
n > e
log(n) = log(e^t)
n = e^t
(a,b) = (0, b) (b!=0)
x = e^t^(1/b)
(a,b)=(1,1)
x = t/W(t)
(a,b)=(2,1)
x = sqrt(2t/W(2t))
(a,b)=(1,2)
x = e^(2W(sqrt(t)/2))
(a,b)=(2,2)
x = e^(W(sqrt(T))
a!=0 b!=0
e^(b/a*W(a/b * t^(1/b))
**/
double lambertWFunction(double z) {
    double w = 5;
    for(int i = 0; i < 15; ++i) {
        double ew = exp(w);
        double den = ew + w*ew;
        if(den < 1e-13)break;
        double x = (w*ew - z) / (ew + w*ew);
        w = w - x;
    }
    return w;
}
double lambertWFunction2(double z) {
    double w = 5;
    for(int i = 0; i < 12; ++i) {
        double ew = exp(w);
        double x = (w*ew - z) / (ew*(w + 1) - (w + 2)*(w*ew - z) / (2 * w + 2));
        w = w - x;
    }
    return w;
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n;
    while(n--) {
        int a, b;
        double t;
        cin >> a >> b >> t;
        double ans = 0;
        if(a == 0) {
            // e^t^(1/b)
            double p = pow(t, 1.0 / b);
            ans = exp(p);
        } else if(b == 0) {
            // n = t^(1/a)
            ans = pow(t, 1.0 / a);
        } else {
            // e^(b/a*W(a/b * t^(1/b))
            double w = lambertWFunction2((double)a / b*pow(t, 1.0 / b));
            ans = exp((double)b / a*w);
        }
        cout << setprecision(20) << ans << endl;
    }
}
 
 
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