1570: Cow Brainiacs 

Time Limit(Common/Java):1000MS/10000MS     Memory Limit:65536KByte
Total Submit: 15            Accepted:8

Description

One afternoon as the cows were chewing their cud, Bessie said, "Let's have a contest to see who is the smartest cow. Here's the contest: we will choose a positive number N (no larger than 2,000,000) and whoever computes the rightmost non-zero digit of N factorial will be crowned the smartest cow."

The other cows demurred, however, mooing, "We did that last year."

"Oh," said Bessie, "but there's a catch. We're not necessarily going to use base 10. I know my hooves don't have that many digits! We'll just specify a positive number base B from 2 through 30."

Write a program to help the cows judge their intelligence contest.

Input

A single line with integers N and B

Output

A single line with the decimal-representation of the "digit" that is the rightmost non-zero digit for N! in base B. If B > 10, go ahead and output a two-digit decimal number as a representation of the final "digit".

Sample Input

Sample Output

Hint

13*12*11*10*9*8*7*6*5*4*3*2*1=6227020800 base 10, which in base 3 is 121001222020102200000, so the right-most non-zero digit is 2.
求n!的k进制下最后一个非0位的数字,我只想喊666
10进制主要有两种,线性模方程的,之后找到考虑hashtable的
这个竟然就是直接暴力的
#include<stdio.h>
int main()
{
int n,b,f=;
scanf("%d%d",&n,&b);
for(int i=;i<=n;i++)
{
f*=i;
while(f%b==)
f/=b;
f%=b;
}
printf("%d",f);
return ;
}

TOJ 假题之 Cow Brainiacs的更多相关文章

  1. HDU 2147--HDU 2147(博弈)--我可能做了假题。。。

    kiki's game Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 40000/1000 K (Java/Others) Total Su ...

  2. 【USACO】Cow Brainiacs

    题意描述 Cow Brainiacs 求 \(n!\) 在 \(b\) 进制表示下的第一位非 \(0\) 位的数字. 算法分析 闲话 忙人自动略过 之前做过一道 \(10\) 进制表示下的题目,感觉差 ...

  3. 杭电15题 The Cow Lexicon

    Problem Description Few know that the cows have their own dictionary with W (1 ≤ W ≤ 600) words, eac ...

  4. Cow Brainiacs

    #include<stdio.h> int main() { ,i; scanf("%d %d",&n,&b); ;i<=n;i++) { f*= ...

  5. TOJ 4008 The Leaf Eaters(容斥定理)

    Description As we all know caterpillars love to eat leaves. Usually, a caterpillar sits on leaf, eat ...

  6. Magic Master(2019年南昌网络赛E题+约瑟夫环)

    目录 题目链接 题意 思路 代码 题目链接 传送门 题意 初始时你有\(n\)张牌(按顺序摆放),每一次操作你将顶端的牌拿出,然后按顺序将上面的\(m\)张牌放到底部. 思路 首先我们发下拿走\(1\ ...

  7. 杭电ACM分类

    杭电ACM分类: 1001 整数求和 水题1002 C语言实验题——两个数比较 水题1003 1.2.3.4.5... 简单题1004 渊子赛马 排序+贪心的方法归并1005 Hero In Maze ...

  8. 转载:hdu 题目分类 (侵删)

    转载:from http://blog.csdn.net/qq_28236309/article/details/47818349 基础题:1000.1001.1004.1005.1008.1012. ...

  9. Codeforces Round #584

    传送门 A. Paint the Numbers 签到. Code #include <bits/stdc++.h> using namespace std; typedef long l ...

随机推荐

  1. chrome浏览器好用的一些插件

    1. Listen 1(听音乐) 2. The Great Suspender(定时释放chrome 打开的页面内存) 3. Similar Web(页面流量分析) 4. Advertising Te ...

  2. css3 变换、过渡效果、动画

    1 CSS3 选择器 1.1 基本选择器 1.2 层级 空格 > + .item+li ~ .item~p 1.3 属性选择器 [attr] [attr=value] [attr^=value] ...

  3. Android 坑爹问题

    A/art: art/runtime/jdwp/jdwp_event.cc:] Check failed: Thread::Current() != GetDebugThread() (Thread: ...

  4. Apache Kafka框架学习

    背景介绍 消息队列的比较 kafka框架介绍 术语解释 文件存储 可靠性保证 高吞吐量实现 负载均衡 应用场景 背景介绍: kafka是由Apache软件基金会维护的一个开源流处理平台,由scala和 ...

  5. sqlserver数据库备份方法

    须事先准备一个工具curl,把它放在c盘.然后,在数据库所在服务器安装7z.最后把这2个存储过程执行,在sqlserver的代理中新建作业,即可实现备份操作. --备份指定数据库到本地和远程指定位置( ...

  6. MFC:AfxLoadLibrary-将指定的 DLL 映射到调用进程的地址空间

    Visual Studio 2012 - Visual C++ LoadLibrary 和 AfxLoadLibrary 进程调用 LoadLibrary (或 AfxLoadLibrary) 以显式 ...

  7. Codeforces Round #277.5 (Div. 2)-D. Unbearable Controversy of Being

    http://codeforces.com/problemset/problem/489/D D. Unbearable Controversy of Being time limit per tes ...

  8. linux虚拟机安装值得注意的几点

    1.建立新的虚拟机时选择自定义安装并选择稍后安装操作系统 2.关键安装命令 tar -xzvf  VMwareTools-10.0.6-3595377.tar.gz sudo ./wmware-ins ...

  9. MacBook Pro休眠掉电、耗电量大问题解决方案

    1.前言 最近我的2015mbpMacBook Pro (Retina, 13-inch, early 2015)更新完10.14系统后,发现休眠待机一晚上后能掉5%电,白天待机4-5小时又掉了8%. ...

  10. ubuntu下如何对接斗鱼直播

    参考教程:https://www.cnblogs.com/liuxuzzz/p/5315998.html 大神写得挺细的,这里都不想再多说了! 为啥要做这个呢?可能真的只是为了好玩吧!!有兴趣直播的孩 ...