C. Voltage Keepsake
You have n devices that you want to use simultaneously.
The i-th device uses ai units of power per second. This usage is continuous. That is, in λ seconds, the device will use λ·ai units of power. The i-th device currently has bi units of power stored. All devices can store an arbitrary amount of power.
You have a single charger that can plug to any single device. The charger will add p units of power per second to a device. This charging is continuous. That is, if you plug in a device for λseconds, it will gain λ·p units of power. You can switch which device is charging at any arbitrary unit of time (including real numbers), and the time it takes to switch is negligible.
You are wondering, what is the maximum amount of time you can use the devices until one of them hits 0 units of power.
If you can use the devices indefinitely, print -1. Otherwise, print the maximum amount of time before any one device hits 0 power.
Input
The first line contains two integers, n and p (1 ≤ n ≤ 100 000, 1 ≤ p ≤ 109) — the number of devices and the power of the charger.
This is followed by n lines which contain two integers each. Line icontains the integers ai and bi (1 ≤ ai, bi ≤ 100 000) — the power of the device and the amount of power stored in the device in the beginning.
Output
If you can use the devices indefinitely, print -1. Otherwise, print the maximum amount of time before any one device hits 0 power.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 4.
Namely, let's assume that your answer is a and the answer of the jury is b. The checker program will consider your answer correct if .
Examples
Input
2 1 2 2 2 1000
Output
2.0000000000
Input
1 100 1 1
Output
-1
Input
3 5 4 3 5 2 6 1
Output
0.5000000000
Note
In sample test 1, you can charge the first device for the entire time until it hits zero power. The second device has enough power to last this time without being charged.
In sample test 2, you can use the device indefinitely.
In sample test 3, we can charge the third device for 2 / 5 of a second, then switch to charge the second device for a 1 / 10 of a second.
题意：给你 n 个用电器的耗电速度和已有的电量，还有一个充电器，已知充电器的充电速度，充电器可以给任何用电器充电，问最后 所有用电器能够同时坚持 的最长时间。
思路：二分答案， 然后通过 check 判断答案是否合理， 合理的条件就是 充电器能够保证所有用电器在这个时间内还没有停止工作，或者 不需要充电器。
AC代码：
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
const double eps = 1e-7;
int a[maxn];
int b[maxn];
int n,p;
bool check(double tim){
double leftim = tim;
for(int i = 0;i < n;i ++){
double tot = a[i] * 1.0 * tim;
double need = tot - b[i] * 1.0;
if(need > p * leftim + eps && need > eps) return 0; /// need > eps 是因为 是直接算的need, need 是有可能小于0的
if(leftim > need / p && need > eps) leftim -= need / p;
}
return 1;
}
int main()
{
while(~scanf("%d%d",&n,&p)){
for(int i = 0;i < n;i ++)
scanf("%d %d",&a[i],&b[i]);
double l = 0.0,r = 1e10;
double t = -1.000;
double flag = 0; /// 标记一下是否可以一直用下去
int T = 100;
while(T --){
double mid = (l + r) / 2;
if(check(mid)){
l = mid;
t = mid;
}
else{
flag = 1;
r = mid;
}
}
if(!flag)
printf("-1\n");
else printf("%.6lf\n",t);
}
return 0;
}