HDU 4588 Count The Carries 计算二进制进位总数

【HDU 4588 Count The Carries 计算二进制进位总数】的更多相关文章

HDU 4588 Count The Carries 计算二进制进位总数

点击打开链接 Count The Carries Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 672 Accepted Submission(s): 230 Problem Description One day, Implus gets interested in binary addition and binary car…

HDU 4588 Count The Carries 数学

Count The CarriesTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87326#problem/C Description One day, Implus gets interested in binary addition and binary carry. He will transfer all decimal digits t…

HDU 4588 Count The Carries（数学统计）

Description One day, Implus gets interested in binary addition and binary carry. He will transfer all decimal digits to binary digits to make the addition. Not as clever as Gauss, to make the addition from a to b, he will add them one by one from a t…

HDU 4588 Count The Carries（找规律，模拟）

题目大意: 求二进制的a加到b的进位数. 思路: 列出前几个2进制,找规律模拟. #include <stdio.h> #include <iostream> #include <algorithm> #include <string.h> #include <math.h> #include <stack> #include <vector> using namespace std; int main() { int…

HDU 4588 Count The Carries （数学，计数）

题意:给定两个十进制数,求二进制中,从x加到y的二进制进了多少位. 析:把这些数字的二进制纵向罗列出来,然后一位一位的把和加起来,最终得到总的进位数.从1到x,第i位上1的总数是x左移i+1位再右移i位后得到的, (在第 0位上,1和0以1010101010的周期出现,并且每个周期一个1,在第1位上,1和0以11001100的周期出现,并且每个周期2个1,以此类推,则第n位上的1的个数是x/2^n*2^(n-1),即先左移n+1位,再右移n位),如果x在i位上面上是1,特殊判断一下,求一下周期以…