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#include <bits/stdc++.h>
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using namespace std;
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const int N = 100010;
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const double eps = 1e-6;
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const double MAX = 1e10;
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//为什么上限要是1e10?浏览了很多题解,都说是浮点数二分题,建议设为1e10
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//卡点:
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//1.MAX设成10^10太小
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//2.eps设成10−8太小导致第19个测试点TLE,提示我们减小精度
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int n;
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double p; //每秒可以给接通的设备充能p个单位
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double a[N]; //第i个设备每秒消耗ai个单位能量。
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double b[N]; //在开始的时候第i个设备里存储着bi个单位能量。
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//检查函数
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//含义:在x这个时长,是否能实现所有设备电能不为零
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bool check(double x) {
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//x: 最多能使用多久
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double sum = p * x; //一共能充多少电
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//遍历每一个设备
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for (int i = 1; i <= n; i++) {
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double t = b[i] - a[i] * x; //设备的原电量,减去,在x这段时间内的消耗电量。
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// 如果大于等于0,表示不需要充电,如果小于0,表示需要充电
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if (t < 0) sum += t; //需要充电t这么多的电量,这些电量需要在sum中扣除掉。
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}
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return sum >= 0; //如果最终sum还是大于等于0的,表示这个x的时间是可以实现的。还可以再大一点试试
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}
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int main() {
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//读入数据
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cin >> n >> p;
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for (int i = 1; i <= n; i++) cin >> a[i] >> b[i];
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//特判(所有设备的每秒耗电量和<=充电宝的每秒充电量,那么是永久性的.输出-1)
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double s = 0;
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for (int i = 1; i <= n; i++) s += a[i];
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if (p >= s) {
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puts("-1");
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exit(0);
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}
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//浮点数二分
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double l = 0, r = MAX;
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while (r - l >= eps) {
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double mid = (l + r) / 2;
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if (check(mid)) l = mid;
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else r = mid;
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}
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printf("%.5f", l);
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return 0;
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}
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