結果

問題 No.456 Millions of Submits!
ユーザー pekempeypekempey
提出日時 2016-12-08 06:45:22
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
WA  
実行時間 -
コード長 3,616 bytes
コンパイル時間 931 ms
コンパイル使用メモリ 84,720 KB
実行使用メモリ 38,912 KB
最終ジャッジ日時 2024-06-23 07:59:27
合計ジャッジ時間 3,538 ms
ジャッジサーバーID
(参考情報)
judge5 / judge4
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 4 ms
5,248 KB
testcase_01 AC 5 ms
5,376 KB
testcase_02 AC 4 ms
5,376 KB
testcase_03 AC 5 ms
5,376 KB
testcase_04 AC 4 ms
5,376 KB
testcase_05 AC 4 ms
5,376 KB
testcase_06 AC 5 ms
5,376 KB
testcase_07 AC 5 ms
5,376 KB
testcase_08 AC 5 ms
5,376 KB
testcase_09 AC 10 ms
5,376 KB
testcase_10 AC 10 ms
5,504 KB
testcase_11 WA -
testcase_12 WA -
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ソースコード

diff #

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cassert>
#include <vector>
#include <algorithm>
using namespace std;

#define getchar getchar_unlocked
#define putchar putchar_unlocked

bool dot;
const int H = 17;
const int N = 1 << H;
const double L = 0;
const double R = log(10);
const double LL = -1e100;
const double RR = log(R);
const double dx = R / (1 << H);
const double logDX = log(R) - H * log(2);
const double log00001 = -4 * log(10);

double lg[N + 1];
bool prime[N + 1];
vector<int> gr[11][11];

int in() {
    int n, c;
    while ((c = getchar()) < '0') {
        assert(c != EOF);
    }
    n = c - '0';
    while ((c = getchar()) >= '0') {
        n = n * 10 + c - '0';
    }
    if (c == '.') {
        dot = true;
    }
    return n;
}

void out(double x) {
    static char buf[20];
    sprintf(buf, "%.9f", x);
    for (int i = 0; buf[i] != '\0'; i++) {
        putchar(buf[i]);
    }
    putchar('\n');
}

void sieve() {
    lg[0] = log(0);
    for (int i = 2; i <= N; i++) {
        if (!prime[i]) {
            double g = log(i);
            for (long long j = i; j < N; j *= i) {
                for (long long k = j; k < N; k += j) {
                    prime[k] = true;
                    lg[k] += g;
                }
            }
        }
    }
    for (int i = 0; i <= N; i++) {
        lg[i] += logDX;
    }
}

int binSearch(int a, int b, double logT) {
    int l = 0;
    int r = N;
    while (r - l > 1) {
        int mid = (l + r) / 2;
        (a * mid * dx + b * lg[mid] >= logT ? r : l) = mid;
    }
    return l;
}

int main() {
    int m = in();
    sieve();
    vector<int> a(m), b(m), c(m), d(m);
    for (int i = 0; i < m; i++) {
        a[i] = in();
        b[i] = in();
        c[i] = in();
        d[i] = dot ? in() : 0;
        c[i] = c[i] * 10000 + d[i];
    }
    vector<int> it(m);
    for (int i = 0; i < m; i++) {
        it[i] = i;
    }
    sort(it.begin(), it.end(), [&](int i, int j) {
        return c[i] < c[j];
    });
    for (int i : it) {
        gr[a[i]][b[i]].emplace_back(i);
    }
    vector<double> ans(m);
    for (int a = 0; a <= 10; a++) {
        for (int b = 0; b <= 10; b++) {
            if (gr[a][b].empty()) {
                continue;
            }
            if (b == 0) {
                for (int i : gr[a][b]) {
                    const double logT = lg[c[i]] - logDX + log00001;
                    ans[i] = exp(logT / a);
                }
                continue;
            }
            if (a == 0) {
                for (int i : gr[a][b]) {
                    const double logT = lg[c[i]] - logDX + log00001;
                    ans[i] = exp(exp(logT / b));
                }
                continue;
            }
            if (gr[a][b].size() <= 500) {
                for (int i : gr[a][b]) {
                    const double logT = lg[c[i]] - logDX + log00001;
                    int l = binSearch(a, b, logT);
                    double s = (lg[l + 1] - lg[l]) / dx;
                    ans[i] = exp((logT + s * b * l * dx - lg[l] * b) / (s * b + a));
                }
            } else {
                int l = 1;
                for (int i : gr[a][b]) {
                    const double logT = lg[c[i]] - logDX + log00001;
                    while (l < N && a * l * dx + b * lg[l] < logT) {
                        l++;
                    }
                    double s = (lg[l + 1] - lg[l]) / dx;
                    ans[i] = exp((logT + s * b * l * dx - lg[l] * b) / (s * b + a));
                }
            }
        }
    }
    for (int i = 0; i < m; i++) {
        out(ans[i]);
    }
}
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