HDOJ 5475 An easy problem

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HDOJ 5475 An easy problem

题目传送门题意:一个计算器,两种操作,乘上x,或者除掉之前的某个x,结果取模输出分析:因为取模不支持除法,然后比赛时想到用逆元,结果发现MOD需要与b互质,结果一直苦苦寻找求逆元的其它方法.后来队友用暴力方法竟然水过,具体操作是记录每次乘的x,如果除的话,将对应的x 改为1,然后一个一个乘.当然正解应该用线段树,树的底部每个点表示每一次操作的x,pushup的是区间的乘积,如果是除把对应的x变为1,发现其实就是暴力的优化.... /******************************…

hdu 5475 An easy problem（暴力 || 线段树区间单点更新）

http://acm.hdu.edu.cn/showproblem.php?pid=5475 An easy problem Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 755 Accepted Submission(s): 431 Problem Description One day, a useless calculator was…

HDU 5475 An easy problem 线段树

An easy problem Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5475 Description One day, a useless calculator was being built by Kuros. Let's assume that number X is showed on the screen of calculator. At first,…

HDOJ 2055 An easy problem

Problem Description we define f(A) = 1, f(a) = -1, f(B) = 2, f(b) = -2, - f(Z) = 26, f(z) = -26; Give you a letter x and a number y , you should output the result of y+f(x). Input On the first line, contains a number T.then T lines follow, each line…

2015上海网络赛 HDU 5475 An easy problem 线段树

题意就不说了思路:线段树,维护区间乘积.2操作就将要除的点更新为1. #include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<queue> #include<vector> #include<set> #include<string> #define inf…

hud-5475 An easy problem(线段树)

http://acm.hdu.edu.cn/showproblem.php?pid=5475 An easy problem Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1146 Accepted Submission(s): 560 Problem Description One day, a useless calculat…

HDOJ(HDU) 2132 An easy problem

Problem Description We once did a lot of recursional problem . I think some of them is easy for you and some if hard for you. Now there is a very easy problem . I think you can AC it. We can define sum(n) as follow: if i can be divided exactly by 3 s…