G(x) Time Limit: 2000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 184    Accepted Submission(s): 44 Problem Description For a binary number x with n digits (AnAn-1An-2 ... A2A1), we encode it as Where ""…

F(x) Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 382    Accepted Submission(s): 137 Problem Description For a decimal number x with n digits (AnAn-1An-2 ... A2A1), we define its weight as F(x…

A Bit Fun Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 423    Accepted Submission(s): 270 Problem Description There are n numbers in a array, as a0, a1 ... , an-1, and another number m. We de…

Minimum palindrome Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 260    Accepted Submission(s): 127 Problem Description Setting password is very important, especially when you have so many "in…

We Love MOE Girls Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 208    Accepted Submission(s): 157 Problem Description Chikami Nanako is a girl living in many different parallel worlds. In this…

Flyer Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 386    Accepted Submission(s): 127 Problem Description The new semester begins! Different kinds of student societies are all trying to adver…

Mex Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 623    Accepted Submission(s): 209 Problem Description Mex is a function on a set of integers, which is universally used for impartial game t…

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=4734 数位DP. 用dp[i][j][k] 表示第i位用j时f(x)=k的时候的个数,然后需要预处理下小于k的和,然后就很容易想了 dp[i+1][j][k+(1<<i)]=dp[i][j1][k];(0<=j1<=j1) AC代码: #include <iostream> #include <cstdio> #include <cstring> #i…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4737 题目大意:给定一系列数,F(i,j)表示对从ai到aj连续求或运算,(i<=j)求F(i,j)<=m的总数. 解题思路:或运算只会让值变大或保持不变.不断通过右移j来更新F(i,j),当aj>=m时所有的i<=j F(i,j)都大于等于m,因此从j后面继续扫数组:当aj<m而F(i,j)>=m时通过右移i来使F(i,j)<m.扫完整个数组即可. #include…

A Bit Fun Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1148    Accepted Submission(s): 644 Problem Description There are n numbers in a array, as a0, a1 ... , an-1, and another number m. We d…