Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1) A
Description
Bear Limak wants to become the largest of bears, or at least to become larger than his brother Bob.
Right now, Limak and Bob weigh a and b respectively. It's guaranteed that Limak's weight is smaller than or equal to his brother's weight.
Limak eats a lot and his weight is tripled after every year, while Bob's weight is doubled after every year.
After how many full years will Limak become strictly larger (strictly heavier) than Bob?
The only line of the input contains two integers a and b (1 ≤ a ≤ b ≤ 10) — the weight of Limak and the weight of Bob respectively.
Print one integer, denoting the integer number of years after which Limak will become strictly larger than Bob.
4 7
2
4 9
3
1 1
1
In the first sample, Limak weighs 4 and Bob weighs 7 initially. After one year their weights are 4·3 = 12 and 7·2 = 14 respectively (one weight is tripled while the other one is doubled). Limak isn't larger than Bob yet. After the second year weights are 36 and 28, so the first weight is greater than the second one. Limak became larger than Bob after two years so you should print 2.
In the second sample, Limak's and Bob's weights in next years are: 12 and 18, then 36 and 36, and finally 108 and 72 (after three years). The answer is 3. Remember that Limak wants to be larger than Bob and he won't be satisfied with equal weights.
In the third sample, Limak becomes larger than Bob after the first year. Their weights will be 3 and 2 then.
题意:一个人每年是*3的增长,一个人是每年*2的增长,问几年后可以超过
解法:模拟
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int main()
{
int i;
int a,b;
cin>>a>>b;
for(i=;a<=b;i++)
{
a*=;
b*=;
}
cout<<i-<<endl;
return ;
}
Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1) A的更多相关文章
- Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1) 菜鸡只会ABC!
Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1) 全场题解 菜鸡只会A+B+C,呈上题解: A. Bear and ...
- Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1) C. Bear and Different Names 贪心
C. Bear and Different Names 题目连接: http://codeforces.com/contest/791/problem/C Description In the arm ...
- Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1) B - Bear and Friendship Condition 水题
B. Bear and Friendship Condition 题目连接: http://codeforces.com/contest/791/problem/B Description Bear ...
- 【树形dp】Codeforces Round #405 (rated, Div. 1, based on VK Cup 2017 Round 1) B. Bear and Tree Jumps
我们要统计的答案是sigma([L/K]),L为路径的长度,中括号表示上取整. [L/K]化简一下就是(L+f(L,K))/K,f(L,K)表示长度为L的路径要想达到K的整数倍,还要加上多少. 于是, ...
- Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1)
A 模拟 B 发现对于每个连通块,只有为完全图才成立,然后就dfs C 构造 想了20分钟才会,一开始想偏了,以为要利用相邻NO YES的关系再枚举,其实不难.. 考虑对于顺序枚举每一个NO/YES, ...
- Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1)A B C 水 并查集 思路
A. Bear and Big Brother time limit per test 1 second memory limit per test 256 megabytes input stand ...
- 【构造】Codeforces Round #405 (rated, Div. 1, based on VK Cup 2017 Round 1) A. Bear and Different Names
如果某个位置i是Y,直接直到i+m-1为止填上新的数字. 如果是N,直接把a[i+m-1]填和a[i]相同即可,这样不影响其他段的答案. 当然如果前面没有过Y的话,都填上0就行了. #include& ...
- Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1) E
Description Bear Limak prepares problems for a programming competition. Of course, it would be unpro ...
- Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1) D
Description A tree is an undirected connected graph without cycles. The distance between two vertice ...
- Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1) C
Description In the army, it isn't easy to form a group of soldiers that will be effective on the bat ...
随机推荐
- ditaa - 把ascii图形转成图片
ditaa ditaa是一个把ascii图形转成图片的工具. 在查看zguide时看到这个文档是用gitdown生成的.zguide文档格式排版非常不错,以后要抽时间好好学习一下. 每章写一个txt文 ...
- u-boot简单学习笔记(三)——AR9331 uboot启动分析
1.最开始系统上电后 ENTRY(_start)程序入口点是 _start 由board/ap121/u-boot.lds引导 2._start: cpu/mips/start.S 是第一个源程序文 ...
- LeetCode(66)题解: Plus One
https://leetcode.com/problems/plus-one/ 题目: Given a non-negative number represented as an array of d ...
- HDOJ--1869--六度分离(用三种算法写的,希望能比較出来他们之间的差别)
六度分离 Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submi ...
- jQuery.ajaxSetup()
jQuery.ajaxSetup()函数用于设置AJAX的全局默认设置. 该函数用于更改jQuery中AJAX请求的默认设置选项.之后执行的所有AJAX请求,如果对应的选项参数没有设置,将使用更改后的 ...
- 如何解决Windows的端口占用问题?
已知某应用在启动时会创建服务套接字,并将其绑定至端口8888,然而端口8888已被占用,如何解除占用? 以下为解决方案: 在cmd中运行netstat -ano|findstr 8888,其中的参数8 ...
- hdoj 1875 畅通project再续【最小生成树 kruskal && prim】
畅通project再续 Problem Description 相信大家都听说一个"百岛湖"的地方吧,百岛湖的居民生活在不同的小岛中,当他们想去其它的小岛时都要通过划小船来实现. ...
- Linux Linker Script
先推荐两个网页: http://blog.csdn.net/muyuyuzhong/article/details/7755291 http://www.cnblogs.com/liulipeng/a ...
- Spring Boot 整合Servlet
冷知识,几乎用不到 在spring boot中使用Servlet有两种实现方法: 方法一: 正常创建servlet,然后只用注解@ServletComponentScan package clc.us ...
- HDU 1081:To The Max
To The Max Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total ...