Pagodas Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 70    Accepted Submission(s): 62 Problem Description n pagodas were standing erect in Hong Jue Si between the Niushou Mountain and the Yun…

Pagodas Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 24    Accepted Submission(s): 22 Problem Description n pagodas were standing erect in Hong Jue Si between the Niushou Mountain and the Yun…

Pagodas Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1282    Accepted Submission(s): 902 Problem Description n pagodas were standing erect in Hong Jue Si between the Niushou Mountain and the…

n pagodas were standing erect in Hong Jue Si between the Niushou Mountain and the Yuntai Mountain, labelled from 1 to n. However, only two of them (labelled a and b, where 1≤a≠b≤n) withstood the test of time. Two monks, Yuwgna and Iaka, decide to mak…

题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=5512 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Problem Descriptionn pagodas were standing erect in Hong Jue Si between the Niushou Mountain and the Yuntai Mou…

nn pagodas were standing erect in Hong Jue Si between the Niushou Mountain and the Yuntai Mountain, labelled from 11 to nn. However, only two of them (labelled aa and bb, where 1≤a≠b≤n1≤a≠b≤n) withstood the test of time. Two monks, Yuwgna and Iaka, d…

题目:传送门. 题意:t组数据,每组数据给定n,a,b,a!=b,在[1,n]的这些点中,每次选取a+b或a-b或b-a点,选取过的点在下次选取的时候可以当做ab来用继续选取,谁不能继续选取谁就输,问最后谁能赢. 题解:首先第一眼看这道题可能会想到博弈,然而这道题本质并不是博弈,而是gcd,因为选取的点一定是 gcd(a,b) 的倍数,可选取的点就是n/gcd(a,b)-2个,所以判断可选取点的奇偶性即可,如果是奇数那么先手赢,否则后手赢. #include <bits/stdc++.h> u…

题目链接 题意:开始有a,b两点,之后可以按照a-b,a+b的方法生成[1,n]中没有的点,Yuwgna 为先手, Iaka后手.最后不能再生成点的一方输: (1 <= n <= 20000) T组数据T <= 500; 思路:由扩展欧几里得知道对于任意正整数,一定存在整数x,y使得 x*a + y*b = gcd(a,b);并且这个gcd是a,b组成的最小正整数:同时也知道了这也是两个点之间的最小距离: 之后直接求点的个数即可: ps:这道题我竟然想到了组合游戏..明显没有说双方都要用…

2015 ACM / ICPC 沈阳现场赛 D 题 找了一小时规律......发现是个GCD. #include<cstdio> #include<cstring> #include<cmath> #include<queue> #include<algorithm> using namespace std; int n,a,b; int gcd(int a,int b) { int t; while(b) t = a%b,a = b,b = t…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5512 学习菊苣的博客,只粘链接,不粘题目描述了. 题目大意就是给了初始的集合{a, b},然后取集合里的两个元素进行加或者减的操作,生成新的元素.问最后最多能生成多少个元素.问答案的奇偶性. 首先一开始有a, b.那么如果生成了b-a(b>a),自然原来的数同样可以由b-a, a生成(b != 2a). 于是如此反复下去,最后的数必然是可以由两个数p, 2p生成的. 于是所有的数肯定可以表示成xp+…