愉快的补题解2 CodeForces - 879A Borya's Diagnosis 大大大大大大模拟！

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本文介绍了一个关于就医安排的问题，病人需按顺序访问多位具有特殊工作时间规律的医生。文章提供了一种模拟解决方案，通过调整访问日期来确保按最优顺序完成所有预约。

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It seems that Borya is seriously sick. He is going visit n doctors to find out the exact diagnosis. Each of the doctors needs the information about all previous visits, so Borya has to visit them in the prescribed order (i.e. Borya should first visit doctor 1, then doctor 2, then doctor 3 and so on). Borya will get the information about his health from the last doctor.

Doctors have a strange working schedule. The doctor i goes to work on the si-th day and works every di day. So, he works on days si, si + di, si + 2di, …

The doctor’s appointment takes quite a long time, so Borya can not see more than one doctor per day. What is the minimum time he needs to visit all doctors?

Input
First line contains an integer n — number of doctors (1 ≤ n ≤ 1000).

Next n lines contain two numbers si and di (1 ≤ si, di ≤ 1000).

Output
Output a single integer — the minimum day at which Borya can visit the last doctor.

Examples
Input
3
2 2
1 2
2 2
Output
4
Input
2
10 1
6 5
Output
11
Note
In the first sample case, Borya can visit all doctors on days 2, 3 and 4.

In the second sample case, Borya can visit all doctors on days 10 and 11.

题目大意：有一群医生都是神经病，他们上班的日期成等差数列，即s,s+d,s+2d…s+k*d，但是你不得不每个医生都找一遍而且每天只能看一个，你说你气不气，气的话就找一个最小的时间段，快点看完。
思路：第一反应肯定是直接模拟，但是仔细想了想应该能推出个公式什么的，就不在这题上浪费时间了，过了算了。。。模拟解法，如果以后挖坟能挖到这道题就再补别的解法。

AC代码：

#include <iostream>

using namespace std;
int main(){
    int n, x, y;
    while(cin>>n){
    	int d = 0;
	    for(int i = 0; i < n; i++){
	       cin>>x>>y;
	       if(d < x) d = x;
	       else{
	         while(x <= d)
			 	x += y;
	         d = x;
	       }
	    }
   	 	cout<<d<<endl;
   	 }
    return 0;
}