# ZOJ 3600 Taxi Fare（数学啊）

Last September, Hangzhou raised the taxi fares.

The original flag-down fare in Hangzhou was 10 yuan, plusing 2 yuan per kilometer after the first 3km and 3 yuan per kilometer after 10km. The waiting fee was 2 yuan per five minutes.
Passengers need to pay extra 1 yuan as the fuel surcharge.

According to new prices, the flag-down fare is 11 yuan, while passengers pay 2.5 yuan per kilometer after the first 3 kilometers, and 3.75 yuan per kilometer after 10km. The waiting fee
is 2.5 yuan per four minutes.

The actual fare is rounded to the nearest yuan, and halfway cases are rounded up. How much more money does it cost to take a taxi if the distance is d kilometers and the waiting
time is t minutes.

#### Input

There are multiple test cases. The first line of input is an integer T ≈ 10000 indicating the number of test cases.

Each test case contains two integers 1 ≤ d ≤ 1000 and 0 ≤ t ≤ 300.

#### Output

For each test case, output the answer as an integer.

```4
2 0
5 2
7 3
11 4
```

#### Sample Output

```0
1
3
5
```

Author: WU, Zejun

Contest: The 9th Zhejiang Provincial Collegiate Programming Contest

PS:

```#include<cstdio>
#include<cmath>
#include<iostream>
using namespace std;
int main()
{
int T,d,t,a,b;
double sum1,sum2;
while(cin >> T)
{
while(T--)
{
cin >> d >> t;
sum1=11;
sum2=11;
if(d>3&&d<=10)
{
sum1+=(d-3)*2;
sum2+=(d-3)*2.5;
}
else if(d>10)
{
sum1+=7*2+(d-10)*3;
sum2+=7*2.5+(d-10)*3.75;
}
sum1+=(t/5.0)*2;
sum2+=(t/4.0)*2.5;
a=(int)(sum1+0.5);
b=(int)(sum2+0.5);
if(a>b)
cout << a - b << endl;
else
cout << b - a << endl;
}
}
return 0;
}```

## [三分法初探]

[Uva Error Curves]一些二次函数取max的交集还是一个下凸函数.三分即可 #include <bits/stdc++.h> #define maxn 10010 using namespace std; int n, a[maxn], b[maxn], c[maxn]; #define F(i, x) a[i] * x * x + b[i] * x + c[i] inline double cal(double x){ double ret = -1e12; for(int i