题目:输入一个整数,求该整数的二进制表达中有多少个1.例如输入10,由于其二进制表示为1010,有两个1,因此输出2. 假设该整数为i.首先i和1做与运算,判断i的最低位是不是为1.接着把1左移一位得到2,再和i做与运算,就能判断i的次高位是不是1……这样反复左移,每次都能判断i的其中一位是不是1.基于此,我们得到如下代码 int NumberOf1_Solution(int i) { int count = 0; unsigned int flag = 1; while(flag) { if(
昨天突然看到一个算法题:一个正整数a的二进制中0的个数: 话说这是个老题了,直观的算法就每次右移一位,直到0为止:代码就省略了: 仔细想想有更好的方案么? 就是这个题可以转换成一个正整数~a的二进制中1的个数: 求1的个数这个貌似就很熟悉了吧: int num = 0; b = ~a; while(b){ num++; b = b & (b-1); } 是不是容易了许多呢 另外像java和python这种没有unsigned的语言要自己去转 b = ~a & 0x0ffff
Round Numbers Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 8590 Accepted: 3003 Description The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, Paper, Stone' (also known as 'Rock, Paper, Scissors',
#include <stdio.h> #include <stdlib.h> /* 4.统计一个整数对应的二进制数的1的个数. 输入一个整数(可正可负), 输出该整数的二进制包含1的个数, “ctl+ z” 结束. */ int main(){ int number; while (scanf("%d", &number) != EOF){ int num[100] = {0}; int length = 0; if (number < 0) nu
问题: Write a function that takes an unsigned integer and returns the number of ’1' bits it has (also known as the Hamming weight). For example, the 32-bit integer ’11' has binary representation 00000000000000000000000000001011, so the function should