Common Knowledge_快速幂
问题 I: Common Knowledge
时间限制: 1 Sec 内存限制: 64 MB
提交: 9 解决: 8
[提交][状态][讨论版]
题目描述

For some odd reason, the two players cannot see the scoreboards entirely. Alice can only see the lower half of her own scoreboard and the upper half of Bob’s scoreboard. Bob can only see the upper half of his scoreboard and the upper half of Alice’s scoreboard. Here, ‘half’ is meant to
include the horizontal segments in the digits’ centers: they can be seen by both players at all times. For example, if one sees the upper half of an eight, one can conclude that the digit is not a zero.

A pair of n-digit scores is called fully known if both players know both scores (i.e. all 2n digits) by looking at the displays with their restricted vision. The players cannot communicate.
输入
• one line with an integer n (1 ≤ n ≤ 20), where n is the number of digits.
输出
样例输入
10
样例输出
1073741824
#include <iostream>
#include <cstdio>
#include <algorithm> using namespace std; int main()
{
int x=;
long long int mi;
int n;
while(scanf("%d",&n)!=EOF){
mi=x;
if(n==){
printf("1\n");
continue;
}
for(int i=;i<n;){
if(i<n/){
mi*=mi;
i*=;
}else{
mi*=x;
i+=;
}
}
printf("%lld\n",mi);
mi=;
}
return ;
}
Common Knowledge_快速幂的更多相关文章
- 【板子】gcd、exgcd、乘法逆元、快速幂、快速乘、筛素数、快速求逆元、组合数
1.gcd int gcd(int a,int b){ return b?gcd(b,a%b):a; } 2.扩展gcd )extend great common divisor ll exgcd(l ...
- jiulianhuan 快速幂--矩阵快速幂
题目信息: 1471: Jiulianhuan 时间限制: 1 Sec 内存限制: 128 MB 提交: 95 解决: 22 题目描述 For each data set in the input ...
- Codeforces Round #209 (Div. 2)A贪心 B思路 C思路+快速幂
A. Table time limit per test 1 second memory limit per test 256 megabytes input standard input outpu ...
- CodeForces 227E Anniversary (斐波那契的高妙性质+矩阵快速幂)
There are less than 60 years left till the 900-th birthday anniversary of a famous Italian mathemati ...
- BNU 4356 ——A Simple But Difficult Problem——————【快速幂、模运算】
A Simple But Difficult Problem Time Limit: 5000ms Memory Limit: 65536KB 64-bit integer IO format: %l ...
- BNU29139——PvZ once again——————【矩阵快速幂】
PvZ once again Time Limit: 2000ms Memory Limit: 65536KB 64-bit integer IO format: %lld Java cla ...
- codeforces 696C C. PLEASE(概率+快速幂)
题目链接: C. PLEASE time limit per test 1 second memory limit per test 256 megabytes input standard inpu ...
- CodeChef Sereja and LCM(矩阵快速幂)
Sereja and LCM Problem code: SEALCM Submit All Submissions All submissions for this problem ar ...
- 小总结:快速幂+贪心————Bit Mask____UVA 10718 多多去理解去温习哦!
传送门:https://vjudge.net/problem/UVA-10718 Preview: bitstream:a flow of data in binary form. in bit-wi ...
随机推荐
- firefox如何卸载插件plugins和临时文件夹
下载原版的 英文版的 firefox 会看到 openH264 video codec Plugin和microsoft DRM (digit rightcopy manager 数字版权管理)等等插 ...
- 浅谈javascript函数节流
浅谈javascript函数节流 什么是函数节流? 函数节流简单的来说就是不想让该函数在很短的时间内连续被调用,比如我们最常见的是窗口缩放的时候,经常会执行一些其他的操作函数,比如发一个ajax请求等 ...
- soj4271 Love Me, Love My Permutation (DFS)
4271: Love Me, Love My Permutation Description Given a permutation of n: a[0], a[1] ... a[n-1], ( it ...
- boss设计参考的脑图
- Redis学习笔记八:独立功能之二进制位数组
Redis 提供了 setbit.getbit.bitcount.bitop 四个命令用于处理二进制位数组. setbit 命令用于为位数组指定偏移量上的二进制位设置值,偏移量从 0 开始计数. ge ...
- HDU 4791 Alice's Print Service(2013长沙区域赛现场赛A题)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4791 解题报告:打印店提供打印纸张服务,需要收取费用,输入格式是s1 p1 s2 p2 s3 p3.. ...
- iOS开发——UI进阶篇(八)pickerView简单使用,通过storyboard加载控制器,注册界面,通过xib创建控制器,控制器的view创建,导航控制器的基本使用
一.pickerView简单使用 1.UIPickerViewDataSource 这两个方法必须实现 // 返回有多少列 - (NSInteger)numberOfComponentsInPicke ...
- Unity手游之路<二>Java版服务端使用protostuff简化protobuf开发
http://blog.csdn.net/janeky/article/details/17151465 开发一款网络游戏,首先要考虑的是客户端服务端之间用何种编码格式进行通信.之前我们介绍了Unit ...
- Windows 10磁盘占用100%解决办法
开机后磁盘占用高,是因为 windows 10 默认启用了 superfetch 服务. 这个服务的主要功能是加快程序的启动速度.开机以后,系统将那些经常使用的程序,预先从硬盘加载到内存中,这样, ...
- 高性能图片服务器–ZIMG
2011年李彦宏在百度联盟峰会上就提到过互联网的读图时代已经到来1,图片服务早已成为一个互联网应用中占比很大的部分,对图片的处理能力也相应地变成企业和开发者的一项基本技能.需要处理海量图片的典型应用有 ...