King's Cake Problem Description It is the king's birthday before the military parade . The ministers prepared a rectangle cake of size n×m(1≤n,m≤10000) . The king plans to cut the cake himself. But he has a strange habit of cutting cakes. Each time,…

King's Cake Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 736    Accepted Submission(s): 539 Problem Description It is the king's birthday before the military parade . The ministers prepared a…

King's Cake 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5640 Description It is the king's birthday before the military parade . The ministers prepared a rectangle cake of size n×m(1≤n,m≤10000) . The king plans to cut the cake himself. But he has…

题目链接 King's Cake Time Limit: 2000/1000 MS (Java/Others)     Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 79    Accepted Submission(s): 69 Problem Description It is the king's birthday before the military parade . The ministers prepar…

题意: 思路: #include<cstdio> #include<cstdlib> #include<iostream> #include<algorithm> #include<cstring> using namespace std; int n,m,cas,ans; int solve(int n,int m) { int t; if(n<m){t=n; n=m; m=t;} ; int s=n/m+solve(m,n%m); re…

Problem Description   It is the king's birthday before the military parade . The ministers prepared a rectangle cake of size n×m(1≤n,m≤10000) . The king plans to cut the cake himself. But he has a strange habit of cutting cakes. Each time, he will cu…

Problem Description It is the king's birthday before the military parade . The ministers prepared a rectangle cake of size n×m(1≤n,m≤10000) . The king plans to cut the cake himself. But he has a strange habit of cutting cakes. Each time, he will cut…

UVA 10831 - Gerg's Cake 题目链接 题意:说白了就是给定a, p.问有没有存在x^2 % p = a的解 思路:求出勒让德标记.推断假设大于等于0,就是有解,小于0无解 代码: #include <stdio.h> #include <string.h> long long a, p; long long pow_mod(long long x, long long k, long long mod) { long long ans = 1; while (k…

Big Number 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1722 ——每天在线,欢迎留言谈论. 题目大意: 给你两个数 n1,n2 . 然后你有一块蛋糕,提前切好,使得不管来 n1 还是 n2 个人都能够当场平均分配. 求 “提前切好” 的最小蛋糕块数. 知识点: (请无视)公式:N = a + b + gcd(a, b) : 思路: (勿无视)先份成p块,然后再拼到一起,再从原来开始的地方,将蛋糕再分成q份,中间肯定会出现完全重合的块…

题意: 给定长方形,每次从中切去一个最大的正方形,问最终可以得到多少正方形. 分析: 过程类似求gcd,每次减去最小的边即可. 代码: #include <cstdio> #include<algorithm> #include<iostream> using namespace std; int main (void) { int T;cin>>T; int n, m; while(T--){ cin>>n>>m; int cnt…

King's Cake Accepts: 967 Submissions: 1572 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Problem Description It is the king's birthday before the military parade . The ministers prepared a rectangle cake of size n×m…

1001:King's Cake(数论) http://acm.hdu.edu.cn/showproblem.php?pid=5640 这题有点辗转相除的意思.基本没有什么坑点. 代码: #include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <algorithm> #include &l…

King's Cake [思路] 递推 公式:f(n,m)=f(max(m,n-m),min(m,n-m))+1,n>m [代码] #include<cstdio> #include<iostream> using namespace std; int max(int a,int b) { return a>b? a:b; } int min(int a,int b) { return a>b? b:a; } int f(int n,int m) { ; ; }…

King's Cake Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 188    Accepted Submission(s): 159 Problem Description It is the king's birthday before the military parade . The ministers prepared a…

King's Order 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5642 Description After the king's speech , everyone is encouraged. But the war is not over. The king needs to give orders from time to time. But sometimes he can not speak things well. So i…

在Windows 7 SP1 电脑上执行Cake的的例子 http://cakebuild.net/docs/tutorials/getting-started ,运行./Build.ps1 报下面的错误 PS D:\Workshop\GitHub\cakebuildexample> ./build.ps1 -Target Default Preparing to run build script... Join-Path : 无法将参数绑定到参数"Path",因为该参数是空值.…

1087: [SCOI2005]互不侵犯King Time Limit: 10 Sec  Memory Limit: 162 MBSubmit: 3336  Solved: 1936[Submit][Status][Discuss] Description 在N×N的棋盘里面放K个国王,使他们互不攻击,共有多少种摆放方案.国王能攻击到它上下左右,以及左上左下右上右下八个方向上附近的各一个格子,共8个格子. Input 只有一行,包含两个数N,K ( 1 <=N <=9, 0 <= K &…

C. Tennis Championship(递推,斐波那契) 题意:n个人比赛,淘汰制,要求进行比赛双方的胜场数之差小于等于1.问冠军最多能打多少场比赛.题解:因为n太大,感觉是个构造.写写小数据,看看有没有结论. 2 3 4 5 6 7 8 9 10 11 12 (人数) 1 2 2 3 3 3 4 4 4 4 4 (比赛数) 发现比赛数的增长成斐波那契.维护一个前缀和即可. #include <bits/stdc++.h> #define ll long long using names…

又是数论题 Q&A Q:你TM做数论上瘾了吗 A:没办法我数论太差了,得多练(shui)啊 题意 题目描述 已知多项式方程: a0+a1x+a2x^2+..+anx^n=0 求这个方程在[1, m ] 内的整数解(n 和m 均为正整数) 输入输出格式 输入格式: 输入文件名为equation .in. 输入共n + 2 行. 第一行包含2 个整数n .m ,每两个整数之间用一个空格隔开. 接下来的n+1 行每行包含一个整数,依次为a0,a1,a2..an 输出格式: 输出文件名为equation…

题目大意 在N×N的棋盘里面放K个国王,使他们互不攻击,共有多少种摆放方案.国王能攻击到它上下左右,以及左上 左下右上右下八个方向上附近的各一个格子,共8个格子. 思路 首先,搜索可以放弃,因为这是一个计数问题,正解几乎不可能是搜索. 我们考虑这样一个决策过程:对于每一行,我们决定放哪些格子.这个决策过程很明显满足无后效性和最优子结构,同时,根据上一行可以递推出这一行的所有可行方案. 所以,我们考虑使用动态规划. 怎么划分阶段呢?根据我们以上的推理,很显然可以根据行来划分阶段. 怎么转移呢?在转…

King's Quest Time Limit: 15000MS Memory Limit: 65536K Case Time Limit: 2000MS Description Once upon a time there lived a king and he had N sons. And there were N beautiful girls in the kingdom and the king knew about each of his sons which of those g…

一.题目 Description 在N×N的棋盘里面放K个国王,使他们互不攻击,共有多少种摆放方案.国王能攻击到它上.下.左.右,以及左上.左下.右上.右下八个方向上附近的各一个格子,共8个格子. Input 只有一行,包含两个数N,K ( 1 <=N <=9, 0 <= K <= N * N) Output 方案数 Sample Input 3 2 Sample Output 16 原题链接→_→bzoj1087: [SCOI2005]互不侵犯King 二.题目分析 其实我们可以…

~>>_<<~ 咳咳!!!今天写此笔记,以防他日老年痴呆后不会解方程了!!! Begin ! ~1~, 首先呢,就看到了一个 gcd(a,b),这是什么鬼玩意呢?什么鬼玩意并不重要,重要的她代表的含义,其实呢,gcd(a,b)就表示 非负整数 a 和 b(不同时为0) 的最大公约数,(数论概论上说:计算 a 与 b 的最大公因数的更低效方法是我女儿四年级老师教的方法,老师要求学生求出 a 与 b 的所有因数,然后找出同时出现在两个表中的最大数字. YES!A good idea f…

Diophantus of Alexandria Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2269    Accepted Submission(s): 851 Problem Description Diophantus of Alexandria was an egypt mathematician living in Ale…

题目链接:HDU 5355 http://acm.split.hdu.edu.cn/showproblem.php?pid=5355 Problem Description There are m soda and today is their birthday. The 1-st soda has prepared n cakes with size 1,2,…,n. Now 1-st soda wants to divide the cakes into m parts so that th…

我们可以用Monad Reader来实现依赖注入(dependency injection DI or IOC)功能.Scala界中比较常用的不附加任何Framework的依赖注入方式可以说是Cake Pattern了.现在通过Monad Reader可以实现同样功能,两者对比优点各有千秋.所谓依赖注入是指在编程时使用了某个未知实现细节的对象,但依赖注入确保这个对象在这段程序运行时已经实例化.这种需求通常是在大型软件开发时对项目进行模块化分割后虽然模块之间互有依赖,但又可以同步开发.特别是在多人…

Lucy and Lily are twins. Today is their birthday. Mother buys a birthday cake for them. Now we put the cake onto a Descartes coordinate. Its center is at (0;0), and the cake's length of radius is 100. There are 2N (N is a integer, 1 N 50) cherries on…

4522: [Cqoi2016]密钥破解 Time Limit: 10 Sec  Memory Limit: 512 MBSubmit: 290  Solved: 148[Submit][Status][Discuss] Description  一种非对称加密算法的密钥生成过程如下: 1.任选两个不同的质数p,q 2.计算N=pq,r=(p−1)(q−1) 3.选取小于r,且与r互质的整数e 4.计算整数d,使得ed≡1KQ/r 5.二元组(N,e)称为公钥,二元组(N,d)称为私钥 当需要加…

#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; typedef long long int64; ]; int64 A,B,K,pi,pk,ans,ti,q; int64 ksm(int64 x,int64 y){ ) ; ) return x; int64 d=k…

http://acm.hdu.edu.cn/showproblem.php?pid=5072 题意:给出N个数,求有多少个三元组,满足三个数全部两两互质或全部两两不互质. 题解: http://dtyfc.com/acm/980 我看的这个学会的…… 可以先求不满足要求的三元组数量,也就是abc,a和b互质,b和c不互质. 这样就要找这n个数中,和某个数不互质的数的个数. 可以质因数分解+容斥原理,求出和某个数不互质的数的个数(也就是和这个数有相同因数的数的个数). 还要先预处理以某个数x为因子…