Codeforces Round #409 (Div. 2) C. Voltage Keepsake 二分

C. Voltage Keepsake

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 λ seconds, 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 i contains 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.

Source

Codeforces Round #409 (rated, Div. 2, based on VK Cup 2017 Round 2)

My Solution

题意：每个设备初始电量为bi，每秒消耗ai，然后充电器每秒可以给一个设备充电p，问所有设备同时工作的最长时长。

二分

很显然的要用二分来做，然后check函数该怎么写呢?

这里题目中说 You can switch which device is charging at any arbitrary unit of time (including real numbers) ，

这就表示可以整体处理，也就是说，首先把设备按照只用初始电量的工作时间(real number)排序，

然后对于每个mid 进行check的时候，把0~i这些工作时长不到mid的设备进行充电，可以得到总时长 sigma{(mid - val[i].dist) * val[i].a} ,然后与mid*p比较即可。

这里相当于在一个设备需要的时候给它充电eps，然后再给另一个需要的设备充电eps，如果sigma < mid*p，即可使所有设备稳定工作mid秒的时长。

此外，

1、对于p >= sigma{ai}的情况表示必定可以充足，此时时间为INF。

2、对于二分时候的EPS取1e-4即可，一般EPS取题目给出的误差精度即可，可自行计算证明。

然后这里二分的r的极限不大好计算，由于double可以保持15位有效数字以上，所以这里取EPS为1e-a时，r可以取(1e15-a)+1。//这里+1是笔者二分时习惯^_^

复杂度 O(nlogn)

#include 
#include 
#include 
#include 
using namespace std;
typedef long long LL;
const int MAXN = 1e5 + 8;
const double EPS = 1e-4;
const double INF = 1e11 + 1;

struct p{
    int a, b;
    double dist;
}val[MAXN];

inline bool cmp(const p &a, const p &b)
{
    return a.dist < b.dist;
}

int n, p;
inline bool check(const double &x)
{
    double sum = 0;
    int i;
    for(i = 0; i < n; i++){
        if(val[i].dist - x > 0) break;
        sum += (x - val[i].dist) * val[i].a;
    }
    if(x * p > sum) return true;
    else return false;

}


int main()
{
    #ifdef LOCAL
    freopen("c.txt", "r", stdin);
    //freopen("c.out", "w", stdout);
    int T = 4;
    while(T--){
    #endif // LOCAL
    ios::sync_with_stdio(false); cin.tie(0);

    int i;
    LL sum = 0;
    cin >> n >> p;
    for(i = 0; i < n; i++){
        cin >> val[i].a >> val[i].b;
        val[i].dist = val[i].b * 1.0 / val[i].a;
        sum += val[i].a;
    }
    sort(val, val + n, cmp);

    if(p - sum >= 0){
        cout << -1 << endl;
    }
    else{
        double ans = 0, l = 0, r = INF, mid;
        while(l + EPS < r){
            mid = (l + r) / 2;
            if(check(mid)){
                ans = mid;
                l = mid;
            }
            else r = mid;
        }
        cout << fixed << setprecision(10) << ans << endl;
    }


    #ifdef LOCAL
    cout << endl;
    }
    #endif // LOCAL
    return 0;
}

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