AtCoder Regular Contest 062 题解

 

C - AtCoDeerくんと選挙速報 / AtCoDeer and Election Report

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Time Limit: 2 sec / Memory Limit: 256 MB

Score : 300300 points

Problem Statement

AtCoDeer the deer is seeing a quick report of election results on TV. Two candidates are standing for the election: Takahashi and Aoki. The report shows the ratio of the current numbers of votes the two candidates have obtained, but not the actual numbers of votes. AtCoDeer has checked the report NN times, and when he checked it for the ii-th (1≦i≦N)(1≦i≦N) time, the ratio was Ti:AiTi:Ai. It is known that each candidate had at least one vote when he checked the report for the first time.

Find the minimum possible total number of votes obtained by the two candidates when he checked the report for the NN-th time. It can be assumed that the number of votes obtained by each candidate never decreases.

Constraints

1≦N≦10001≦N≦10001≦Ti,Ai≦1000(1≦i≦N)1≦Ti,Ai≦1000(1≦i≦N)TiTi and AiAi (1≦i≦N)(1≦i≦N) are coprime.It is guaranteed that the correct answer is at most 10181018.

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Input

The input is given from Standard Input in the following format:

NN T1T1 A1A1 T2T2 A2A2 :: TNTN ANAN

Output

Print the minimum possible total number of votes obtained by Takahashi and Aoki when AtCoDeer checked the report for the NN-th time.

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Sample Input 1 Copy

Copy

3 2 3 1 1 3 2

Sample Output 1 Copy

Copy

10

When the numbers of votes obtained by the two candidates change as 2,3→3,3→6,42,3→3,3→6,4, the total number of votes at the end is 1010, which is the minimum possible number.

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Sample Input 2 Copy

Copy

4 1 1 1 1 1 5 1 100

Sample Output 2 Copy

Copy

101

It is possible that neither candidate obtained a vote between the moment when he checked the report, and the moment when he checked it for the next time.

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Sample Input 3 Copy

Copy

5 3 10 48 17 31 199 231 23 3 2

Sample Output 3 Copy

Copy

6930

 

题意：

给你n时间的人数比例，算出最后的最小人数。

分析：

一个简单思维题，

a b

a1 b1

只要保证a1*x>=a b1*x>=b即可

import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner input=new Scanner( System.in); long n=input.nextLong(); long a=input.nextLong(); long b=input.nextLong(); for(int i=1;i<n;i++) { // System.out.println("s:"+a+" "+b); long a1=input.nextLong(); long b1=input.nextLong(); if(a1>=a&&b1>=b) { a=a1; b=b1; continue; } long d1=a/a1; if(a