codeforces 490 D Chocolate
题意:给出a1*b1和a2*b2两块巧克力,每次可以将这四个数中的随意一个数乘以1/2或者2/3,前提是要可以被2或者3整除,要求最小的次数让a1*b1=a2*b2,并求出这四个数最后的大小。
做法:非常显然仅仅跟2跟3有关。所以s1=a1*b1,s2=a2*b2,s1/=gcd(s1,s2),s2/=gcd(s1,s2),然后若s1跟s2的质因子都是2跟3,那么就有解。之后暴力乱搞就好了。
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<climits>
#include<list>
#include<iomanip>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b)
{
return b==0?a:gcd(b,a%b);
}
bool work(ll x,int *cnt)
{
while(x%2==0)
{
x/=2;
cnt[2]++;
}
while(x%3==0)
{
x/=3;
cnt[3]++;
}
return x==1;
}
void cg(int &a,int &b,int val,int num)
{
while(a%val==0&&num>0)
{
num--;
a/=val;
if(val==3)
a*=2;
}
while(b%val==0&&num>0)
{
num--;
b/=val;
if(val==3)
b*=2;
}
}
int num[2][4];
int a[2],b[2];
ll s[2];
int main()
{
for(int i=0;i<2;i++)
{
cin>>a[i]>>b[i];
s[i]=ll(a[i])*b[i];
}
ll t=gcd(s[0],s[1]);
s[0]/=t;s[1]/=t;
if(!work(s[0],num[0])||!work(s[1],num[1]))
{
puts("-1");
return 0;
}
int ans=0;
for(int i=3;i>1;i--)
{
int sub=abs(num[0][i]-num[1][i]);
ans+=sub;
if(num[0][i]>num[1][i])
{
num[0][i-1]+=sub;
cg(a[0],b[0],i,sub);
}
else
{
num[1][i-1]+=sub;
cg(a[1],b[1],i,sub);
}
}
printf("%d\n%d %d\n%d %d",ans,a[0],b[0],a[1],b[1]);
}
1 second
256 megabytes
standard input
standard output
Polycarpus likes giving presents to Paraskevi. He has bought two chocolate bars, each of them has the shape of a segmented rectangle. The first bar is a1 × b1 segments
large and the second one is a2 × b2 segments
large.
Polycarpus wants to give Paraskevi one of the bars at the lunch break and eat the other one himself. Besides, he wants to show that Polycarpus's mind and Paraskevi's beauty are equally matched, so the two bars must have the same number of squares.
To make the bars have the same number of squares, Polycarpus eats a little piece of chocolate each minute. Each minute he does the following:
- he either breaks one bar exactly in half (vertically or horizontally) and eats exactly a half of the bar,
- or he chips of exactly one third of a bar (vertically or horizontally) and eats exactly a third of the bar.
In the first case he is left with a half, of the bar and in the second case he is left with two thirds of the bar.
Both variants aren't always possible, and sometimes Polycarpus cannot chip off a half nor a third. For example, if the bar is 16 × 23, then
Polycarpus can chip off a half, but not a third. If the bar is 20 × 18, then Polycarpus can chip off both a half and a third. If the bar is 5 × 7,
then Polycarpus cannot chip off a half nor a third.
What is the minimum number of minutes Polycarpus needs to make two bars consist of the same number of squares? Find not only the required minimum number of minutes, but also the possible sizes of the bars after the process.
The first line of the input contains integers a1, b1 (1 ≤ a1, b1 ≤ 109)
— the initial sizes of the first chocolate bar. The second line of the input contains integers a2, b2 (1 ≤ a2, b2 ≤ 109)
— the initial sizes of the second bar.
You can use the data of type int64 (in Pascal), long
long (in С++), long (in Java) to process large integers (exceeding 231 - 1).
In the first line print m — the sought minimum number of minutes. In the second and third line print the possible sizes of the bars
after they are leveled in m minutes. Print the sizes using the format identical to the input format. Print the sizes (the numbers
in the printed pairs) in any order. The second line must correspond to the first bar and the third line must correspond to the second bar. If there are multiple solutions, print any of them.
If there is no solution, print a single line with integer -1.
2 6
2 3
1
1 6
2 3
36 5
10 16
3
16 5
5 16
3 5
2 1
-1
codeforces 490 D Chocolate的更多相关文章
- Codeforces Problem 598E - Chocolate Bar
Chocolate Bar 题意: 有一个n*m(1<= n,m<=30)的矩形巧克力,每次能横向或者是纵向切,且每次切的花费为所切边长的平方,问你最后得到k个单位巧克力( k <= ...
- Codeforces 633F The Chocolate Spree 树形dp
The Chocolate Spree 对拍拍了半天才知道哪里写错了.. dp[ i ][ j ][ k ]表示在 i 这棵子树中有 j 条链, 是否有链延伸上来. #include<bits/ ...
- Codeforces 617B:Chocolate(思维)
题目链接http://codeforces.com/problemset/problem/617/B 题意 有一个数组,数组中的元素均为0或1 .要求将这个数组分成一些区间,每个区间中的1的个数均为1 ...
- codeforces 598E E. Chocolate Bar(区间dp)
题目链接: E. Chocolate Bar time limit per test 2 seconds memory limit per test 256 megabytes input stand ...
- codeforces 633F The Chocolate Spree (树形dp)
题目链接:http://codeforces.com/problemset/problem/633/F 题解:看起来很像是树形dp其实就是单纯的树上递归,就是挺难想到的. 显然要求最优解肯定是取最大的 ...
- Codeforces 598E:Chocolate Bar
E. Chocolate Bar time limit per test 2 seconds memory limit per test 256 megabytes input standard in ...
- Codeforces 633F - The Chocolate Spree(树形 dp)
Codeforces 题目传送门 & 洛谷题目传送门 看来我这个蒟蒻现在也只配刷刷 *2600 左右的题了/dk 这里提供一个奇奇怪怪的大常数做法. 首先还是考虑分析"两条不相交路径 ...
- Codeforces Round #340 (Div. 2) B. Chocolate 水题
B. Chocolate 题目连接: http://www.codeforces.com/contest/617/problem/D Descriptionww.co Bob loves everyt ...
- Codeforces Round #310 (Div. 1) C. Case of Chocolate set
C. Case of Chocolate Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/555/ ...
随机推荐
- Vue.js和Nodejs的关系
首先vue.js 是库,不是框架,不是框架,不是框架. Vue.js 使用了基于 HTML 的模版语法,允许开发者声明式地将 DOM 绑定至底层 Vue 实例的数据. Vue.js 的核心是一个允许你 ...
- NOIP2012D2T1 同余方程
[NOIP2012T4]同余方程 noip2012-tg 描述 求关于 x的同余方程 ax ≡ 1(mod b) 的最小正整数解. 输入格式 输入文件 mod.in 输入只有一行,包含两个正整数a,b ...
- Java单例模式解析(收藏)
在GoF的23种设计模式中,单例模式是比较简单的一种.然而,有时候越是简单的东西越容易出现问题.下面就单例设计模式详细的探讨一下. 所谓单例模式,简单来说,就是在整个应用中保证只有一个类的实例存在.就 ...
- webstorm主题网址
http://www.phpstorm-themes.com/ http://www.riaway.com/theme.php
- centos7 修改默认语言
vi /etc/locale.conf # 修改成英文 LANG="en_US.UTF-8" # 修改成中文 LANG="zh_CN.UTF-8"
- 设置Pycharm在创建py文件时自动添加文件头注释(类似于钩子特性)
在每次新建一个py文件的时候 1 如何自动添加 #!/usr/bin/env python2 自动添加 #-*- coding: utf-8 -*- 操作方法: File -> se ...
- bzoj 1296: [SCOI2009]粉刷匠 动态规划
Description windy有 N 条木板需要被粉刷. 每条木板被分为 M 个格子. 每个格子要被刷成红色或蓝色. windy每次粉刷,只能选择一条木板上一段连续的格子,然后涂上一种颜色. 每个 ...
- loadrunner笔记----好记性不如烂笔头
1.Loadrunner主要由Vugen,Controller和Analyais3部分组成 2.简述描述集合点和集合点函数 集合点可以同步虚拟用户,以便能在同一时刻执行任务,集合点函数lr_rende ...
- python中的全局变量、局部变量、实例变量
1.全局变量:在模块内,在所有函数.类外面. 2.局部变量:在函数内,在类方法内(未加self修饰的) 3.静态变量:在类内,但不在类方法内.[共同类所有,值改变后,之后所有的实例对象也改变] 4.实 ...
- C语言基础 (10) 变量作用域,生命周期 内存结构
01 课程回顾 1.指针数组 注意: 对于数组来说,在使用sizeof的时候a和&a[0]是不一样的, 虽然以%x打印出来他们都是地址 2.值传递 int a; fun(a); int *** ...