高低位交换 题目链接 这道题非常水,我是用位运算做的 a=n>>16 二进制的“高位”b=n-(a<<16) 二进制的“低位”ans=(b<<16)+a 转换 #include<iostream> #include<cstring> #include<cstdio> using namespace std; #define int long long int n; #undef int int main() #define int l
左移右移的能够使得数字*2或者/2 那*3怎么办,就左移一位然后再+ 经典面试题: 1.交换2个数,不用temp a=10 b=12 1.1 a = a + b = 22 b = a - b = 10 a = a - b =10 1.2 位运算 ^(异或) a=1010 b = 1100 a = a ^ b =0110 b = a ^ b =1010 a = a ^ b = 1100
这个题简单来说就是把一个数转成32位的2进制数,不够的补0.然后把这个数的前半部分和后半部分互换,再计算结果. 思路简单明了,接下来是代码: #include<iostream> #include<cstdio> #include<map> #include<cmath> using namespace std; long long a,b,shu[50],s=1,zshu; int main() { scanf("%lld",&
Dollar Dayz Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 5655 Accepted: 2125 Description Farmer John goes to Dollar Days at The Cow Store and discovers an unlimited number of tools on sale. During his first visit, the tools are sell
//将n左移m位 int byte_to_left_move(int n , int m) { int i , ret = 1 ; if(n == 0 || n < 0) { return ; } if(n == 1) { for(i = 0 ; i < m ; i++) ret *= n * 2 ; } if(n > 1) { for(i = 0 ; i < m ; i++) ret *= n ; } return ret ; } //将n右移m位 int byte_to_rig
