mowayao的专栏

思路与uva10892类似#include #include #include using namespace std;long long gcd(long long a,long long b){ return b == 0 ?a : gcd(b,a%b);}int main(){ int ncase; vector num; cin >> ncase; while(n

BeavergnawTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6204 Accepted: 4090DescriptionWhen chomping a tree the beaver cuts a very specific shape out o

Revenge of FibonacciTime Limit: 10000/5000 MS (Java/Others)    Memory Limit: 204800/204800 K (Java/Others)Total Submission(s): 1944    Accepted Submission(s): 446Problem DescriptionThe wel

Harry Potter and the Hide StoryTime Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2193    Accepted Submission(s): 530Problem Descripti

StoneTime Limit: 3000/2000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 2267    Accepted Submission(s): 568Problem DescriptionGiven an array of int

FishnetTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 1788 Accepted: 1144DescriptionA fisherman named Etadokah awoke in a very small island. He could s

Grandpa's EstateTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 10851 Accepted: 2953DescriptionBeing the only living descendant of his grandfather, Kamr

GCD of SequenceTime Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 949    Accepted Submission(s): 284Problem DescriptionAlice is pla

A Round Peg in a Ground HoleTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 5301 Accepted: 1662DescriptionThe DIY Furniture company specializes in assem

Most Distant Point from the SeaTime Limit: 5000MS Memory Limit: 65536KTotal Submissions: 3955 Accepted: 1847 Special JudgeDescriptionThe main land of Japan ca

DP?Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 128000/128000 K (Java/Others)Total Submission(s): 1930    Accepted Submission(s): 640Problem DescriptionFigure 1 shows t

EllipsoidTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1015    Accepted Submission(s): 359Special JudgeProblem DescriptionGive

Help!Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 180    Accepted Submission(s): 32Problem Description“Help! Help!”While

HDU 5033 类凸包

SawtoothTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 979    Accepted Submission(s): 375Problem DescriptionThink about a plane

题意： 问 gcd(i,j) = i ^ j  的对数（j 思路：容易想到  形如  （2,3） （4,5）.....这种互质相邻且二进制位数相同的数一定满足要求。那么对于gcd为2情况进行分析：从gcd(a,b) = 2得到a/2,b/2互质，可以想到a/2与b/2相差只能是1，因为要使a^b = 2 a,b只有在第1位有差别，即差别为2，如果a/2与b/2相差超过1，那么a,b就不

How many integers can you findTime Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3867    Accepted Submission(s): 1088Problem Descripti

定理如下：对任意非负整数a和任意正整数b, gcd(a,b） = gcd(b,a mod b)首先证明 gcd(a,b) | gcd(b,a mod b)设 gcd(a,b) = da mod b = a - b*k (k = a/b 向下取整的整数)易得 d | a mod b 和 d | b 得出 d | gcd(b,a mod b) (d 为 最大公约数的一个因数)接

The ProblemFrom Euclid it is known that for any positive integers A and B there exist such integers X and Y that AX+BY=D, where D is the greatest common divisor of A and B. The problem is to find

基础的数论题：对n进行因式分解，因为两个数的LCM一定是LCM的因数。根据公式a*b = lcm(a,b)*gcd(a,b) 判断a,b是否为所求。#include #include #include #include using namespace std;int gcd(int a,int b){ return b == 0 ? a : gcd(b,a%b);}int main

这题搞了1个多小时= =！靠着UVA的BBS上的数据才过的 思路是：把m质因数分解，各质因数的系数用map(a)记录，把1到n的数都质因数分解，用map(b)记录（累加）,然后逐一遍历,碰到a[i]=0则跳过，碰到b[i] = 0& a[i]!= 0 则无解，接下来就是我无法理解地方了，判断前两种情况之后，如果碰到b[i] 无法整除a[i] 的时候，理应是无解，但是我只要这样判 就是WA，如果不判

简单的数论题：主要的注意点是计算幂的时候尽量少用pow，精度损失太大。#include #include #include #include #include #include #include #include using namespace std;const int maxn = 100 + 10;bool vis[maxn];map mm;vectorans;vo

主要是公式的推导和快速幂的使用，弹了一次，因为交大出的那本SCL快速幂的模板有点小错误#include #include #include using namespace std;long long pow_mod(long long a,long long i,long long n){ if(i == 0) return 1%n; long long tmp = pow_m

#include #include #include #include using namespace std;int main(){ int ncase,T=1; cin >> ncase; string num; while(ncase--){ cin >> num; int sum = 0; int a=0,b=0,c=0; bool flag = 0;

水题：upper_bound,lower_bound解决问题#include #include #include #include #include using namespace std;int main(){ int a,b; vector squ; vector::iterator low,up; for(int i = 1; i <= 1000; i++){

规律题：被n=2的这种情况坑了！#include #include #include using namespace std;int main(){ vector dd; dd.push_back(0); dd.push_back(0); dd.push_back(1); for(int i = 2; i <= 17000; i++){ int tmp = ((i+1)*

简单的排列组合题：英语差一开始在纠结这个digit的意思，后来知道digit是一位数的意思，做法很简单，就是统计每个数的个数，用map记录，如1112,1的个数是3,2的个数是1。z最后计算的时候除去个数的阶乘即可。#include #include #include #include #include #include using namespace std;#define ULL

Andrew has just made a breakthrough in group theory: he realized that he can classify all finite Abelian groups (not much of a breakthrough, indeed). Given n, how many Abelian groups with n elements

We consider problems concerning the number of ways in which a number can be written as a sum. If the order of the terms in the sum is taken into account the sum is called a composition and the numbe

出处http://www.cnblogs.com/yan-boy/archive/2012/11/29/2795294.html矩阵的快速幂是用来高效地计算矩阵的高次方的。将朴素的o（n）的时间复杂度，降到log（n）。这里先对原理（主要运用了矩阵乘法的结合律）做下简单形象的介绍：一般一个矩阵的n次方，我们会通过连乘n-1次来得到它的n次幂。但做下简单的改进就能

CouponsInput: standard inputOutput: standard outputTime Limit: 2 secondsMemory Limit: 32 MB Coupons in cereal boxes are numbered 1 to n, and a set of one of each is required for a prize (a

Problem F: Colossal Fibonacci Numbers!The i'th Fibonacci number f (i) is recursively defined in the following way:f (0) = 0 and f (1) = 1f (i+2) = f (i+1) + f (i)  for every i ≥ 0Your task i

Problem BYet another Number SequenceInput: standard inputOutput: standard outputTime Limit: 3 secondsLet's define another number sequence, given by the following function:f(0) = af(1

IInteger Game  Two players, S and T, are playing a game where they make alternate moves. S plays first. In this game, they start with an integer N. In each move, a player remov

Problem AThe Fibonacci PrimesInput: standard inputOutput: standard outputTime Limit: 8 secondsMemory Limit: 32 MB The Fibonacci number sequence is 1, 1, 2, 3, 5, 8, 13 and so on. You

UVa 12683 Odd and Even Zeroes（数论+数位DP）