Problem about GCD Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 470    Accepted Submission(s): 77 Problem Description Given integer m. Find multiplication of all 1<=a<=m such gcd(a, m)=1 (cop…

普通NIM规则加上一条可以分解为两堆,标准的Multi-SG游戏 一般Multi-SG就是根据拓扑图计算SG函数,这题打表后还能发现规律 sg(1)=1 sg(2)=2 sg(3)=mex{0,1,2,1^2}=4 sg(4)=mex{0,1,2,sg(3)}=3 可以发现3和4的时候相当于互换了位置 /** @Date : 2017-10-12 21:20:21 * @FileName: HDU 3032 博弈 SG函数找规律.cpp * @Platform: Windows * @Autho…

数据范围很大,用米勒罗宾测试和Pollard_Rho法可以分解大数. 模板在代码中 O.O #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> using namespace std; __int64 pri[]= {,,,,,,,,,,};//用小素数表做随机种子避免第一类卡米歇尔数的误判 __int64 mul…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1030 Delta-wave Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7163    Accepted Submission(s): 2772 Problem Description A triangle field is numbe…

A Boring Question Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 865    Accepted Submission(s): 534 Problem Description There are an equation.∑0≤k1,k2,⋯km≤n∏1⩽j<m(kj+1kj)%1000000007=?We define…

Starting from point (0,0) on a plane, we have written all non-negative integers 0, 1, 2,... as shown in the figure. For example, 1, 2, and 3 has been written at points (1,1), (2,0), and (3, 1) respectively and this pattern has continued. You are to w…

已知有n个单位的水,问有几种方式把这些水喝完,每天至少喝1个单位的水,而且每天喝的水的单位为整数.看上去挺复杂要跑循环,但其实上,列举几种情况之后就会发现是找规律的题了= =都是2的n-1次方,而且这题输出二进制数就行了......那就更简单了,直接输出1,然后后面跟n-1个0就行了╮(╯_╰)╭ 下面AC代码 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm>…

题目 //找规律,123321123321123321…发现这样排列恰好可以错开 // 其中注意题中数据范围: M是行,N是列,3 <= N < 2×M //则猜测:m,m,m-1,m-1,m-2,m-2,……,2,2,1,1求出前m个数字的和就是答案. //发现案例符合(之前的代码第二天发现案例都跑不对,真不知道我当时眼睛怎么了) #include <iostream> #include<stdio.h> #include<string.h> #inclu…

主要还是找规律,然后大数相乘 #include<stdio.h> #include<string.h> #include<math.h> #include<time.h> #include<map> #include<iostream> #include<ctype.h> #include<string> #include<algorithm> #include<stdlib.h> #i…

M=1:aaaaaaaa…… M=2:DFS+manacher, 暴出N=1~25的最优解,找规律.N<=8的时候直接输出,N>8时,头两个字母一定是aa,剩下的以aababb循环,最后剩余<5全部补a,等于5补aabab. M=3:abcabcabcabc…… #include <cstdio> #include <cstring> using namespace std; const char str[] = "aababb"; int m…