Winter holiday is coming!As the monitor, LeLe plans to go to the movies.
Because the winter holiday tickets are pretty expensive, LeLe decideds to try group-buying.

There are multiple test cases, about 20 cases. The first line of input contains two integers n,m(1≤n,m≤100). n indicates the number of the students. m indicates how many cinemas have offered group-buying.

For the m lines,each line contains two integers ai,bi(1≤ai,bi≤100), indicating the choices of the group buying cinemas offered which means you can use bi yuan to buy ai tickets in this cinema.

For each case, please help LeLe **choose a cinema** which costs the least money. Output the total money LeLe should pay.

3 2
2 2
3 5

4

Hint
LeLe can buy four tickets with four yuan in cinema 1.
 

BestCoder Round #34

hujie   |   We have carefully selected several similar problems for you:  5193 5192 5191 5189 5188 

BestCoder官方答案：

1001 Go to movie

直接枚举去哪个电影院，从中取花费最少的，ans=min(n+ai?1ai?bi)

水题......直接枚举就行，对于每一家电影院。用n个学生除以团购的学生数，如果有余数多加b[i]块钱，否则就是正好a[i]*b[i]块钱。

import java.io.*;
import java.util.*;

public class Main
{

	public static void main(String[] args)
	{
		// TODO Auto-generated method stub
		Scanner input = new Scanner(System.in);
		while(input.hasNext())
		{
			int n = input.nextInt();  //n表示参见学生的数量
			int m = input.nextInt();  //m表示团购的电影票数
			int a[] = new int[m];
			int b[] = new int[m];
			int c[] = new int[m];     //c数组记录
			for(int i = 0;i<m;i++)
			{
				a[i]=input.nextInt();
				b[i]=input.nextInt();
				int temp = n/a[i];
				if(n%a[i]>0)
				{
					c[i]=(temp+1)*b[i];
				}
				else 
				{
					c[i]=temp*b[i];
				}
			}
			Arrays.sort(c);
			System.out.println(c[0]);
		}
	}

}

寒假啦！为了打发无聊的时光，同时增进男女之间的感情，身为班长的乐乐打算带领大家看电影。
寒假时期的电影票往往很贵，于是乐乐决定团购。

有多组测试数据，大约20组。
输入包含多组数据，对于每组数据，第一行包含两个整数n和m，n表示有多少学生参加，m表示提供团购的电影院数。
接下来m行，每行两个整数a和b，表示每个电影院提供的团购方案，可以用b元买a张票。(1≤n,m,a,b≤100)

对于每组数据，选择一个电影院，输出最少的花费。

3 2
2 2
3 5

4

样例解释，可以团购两次电影院1，花费4.