校内训练2019-11-15逮虾户

Posted 2020-11-23 chhokmah

【思路要点】

• (t=sum_{i=1}^{n} {frac{s_i}{d+v_i}})
• 因为保证了题目的速度为正数，所以会发现这是一个减函数。
• 就用二分逼近答案。

【代码】

#include <bits/stdc++.h>

#define REP(i, s, t) for (int i = s; i <= t; i++)
#define PER(i, s, t) for (int i = s; i >= t; i--)
#define FI first
#define SE second
#define mp make_pair
#define pb push_back

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;

template <class T>
void chkmax(T& x, T y) {
    x = max(x, y);
}
template <class T>
void chkmin(T& x, T y) {
    x = min(x, y);
}

template <class T>
void re(T& x) {
    x = 0;
    char ch = 0;
    int f = 1;
    for (; !isdigit(ch); ch = getchar())
        if (ch == '-')
            f = -1;
    for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
    x *= f;
}

const int N = 1005;
const double eps = 1e-8;

int n, t;
int s[N], v[N];

double check(double d) {
    double ans = 0;
    for (int i = 1; i <= n; i++) ans += s[i] / (d + v[i]);
    return ans;
}

int main() {
    freopen("dejavu.in", "r", stdin);
    freopen("dejavu.out", "w", stdout);
    re(n);
    re(t);
    double l = -1e9, r = 1e9;
    for (int i = 1; i <= n; i++) {
        re(s[i]), re(v[i]);
        l = max(l, -1.0 * v[i]);
    }
    while (r - l > eps) {
        double mid = (l + r) / 2.0;
        if (check(mid) >= t)
            l = mid;
        else
            r = mid;
    }
    printf("%.9f
", l);
    return 0;
}