Codeforces 1176A Divide it!
题目链接:http://codeforces.com/problemset/problem/1176/A

思路:贪心,对第二个操作进行俩次等于将n变成n/3,第三个操作同理,我们将n不断除以2,再除以3,最后除以5,判断最后是否等于1即可。
AC代码:
#include<iostream>
using namespace std;
long long n,ans;
int main(){
int T;
cin >> T;
while(T--){
cin >> n;
ans = ;
while(n % == )n /= ,ans++;
while(n % == )n /= ,ans += ;
while(n % == )n /= ,ans += ;
if(n == ) cout << ans << endl;
else cout << - << endl;
}
return ;
}
Codeforces 1176A Divide it!的更多相关文章
- CodeForces - 1176A Divide it! (模拟+分类处理)
You are given an integer nn. You can perform any of the following operations with this number an arb ...
- codeforces 792C. Divide by Three
题目链接:codeforces 792C. Divide by Three 今天队友翻了个大神的代码来问,我又想了遍这题,感觉很好,这代码除了有点长,思路还是清晰易懂,我就加点注释存一下...分类吧. ...
- A - Divide it! CodeForces - 1176A
题目: You are given an integer nn. You can perform any of the following operations with this number an ...
- CodeForces - 792C Divide by Three (DP做法)
C. Divide by Three time limit per test: 1 second memory limit per test: 256 megabytes input: standar ...
- Codeforces 977D Divide by three, multiply by two(拓扑排序)
Polycarp likes to play with numbers. He takes some integer number xx, writes it down on the board, ...
- Codeforces 1270E - Divide Points(构造+奇偶性)
Codeforces 题目传送门 & 洛谷题目传送门 显然,直接暴力枚举是不可能的. 考虑将点按横纵坐标奇偶性分组,记 \(S_{i,j}=\{t|x_t\equiv i\pmod{2},y_ ...
- CodeForces 792C - Divide by Three [ 分类讨论 ]
删除最少的数位和前缀0,使得剩下的数能被3整除 等价于各数位数字之和能被3整除. 当前数位和可能是 0, 1, 2(mod 3) 0: 直接处理 1: 删除一个a[i]%3 == 1 或者 两个a[i ...
- 3.26-3.31【cf补题+其他】
计蒜客)翻硬币 //暴力匹配 #include<cstdio> #include<cstring> #define CLR(a, b) memset((a), (b), s ...
- Divide by three, multiply by two CodeForces - 977D (思维排序)
Polycarp likes to play with numbers. He takes some integer number xx, writes it down on the board, a ...
随机推荐
- RVIZ可视化平台
- 自己写的一些Delphi常用函数
今天在整理以前写过的代码,发现有些函数还是挺实用的,决定将其贴到Blog上,与众多好友一起分享.{*************************************************** ...
- JAVA中HashMap相关知识的总结(一)
Java中HashMap在jdk1.7和jdk1.8中的区别点: 在jdk1.7中是用数组+链表形式存储,1.8采用数组+链表/红黑树形式 Jdk1.8中由链表转为红黑树是长度大于8,由红黑树转为链表 ...
- Codeforces 1189A Keanu Reeves
题目链接:http://codeforces.com/problemset/problem/1189/A 思路:统计1 和 0 的个数,不相等拆开字符串,否则不拆. AC代码: #include< ...
- python 实现异常退出
https://blog.csdn.net/u013385362/article/details/81206822 有时当一个条件成立的情况下,需要终止程序,可以使用sys.exit()退出程序.sy ...
- shell 删除超过30天的文件和目录
#!/bin/bash location="/root/sqlbak/" find $location -mtime +30 -type d |xargs rm -rf #删除目录 ...
- Java文件系统
Java7 引入了新的输入/输出2(NIO.2)API并提供了一个新的I/O API. 它向Java类库添加了三个包:java.nio.file,java.nio.file.attribute和jav ...
- hadoop备战:hbase的分布式安装经验
配置HBase时,首先考虑的肯定是Hbase版本号与你所装的hadoop版本号是否匹配.这一点我在之前 的博客中已经说明清楚,hadoop版本号与hbase版本号的匹配度,那是官方提供的.以下的实验就 ...
- css3条纹进度条
新建div,取名progress,如下 <div class="progress"></div> 在里面插入条纹进度条,以及进度显示文本进度: <di ...
- 个人使用Viso绘制的简单神经网络实现原理图