poj_3748_位操作.cpp

Problem Links:

poj3748,

Problem:

位操作

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5792		Accepted: 2219

Description

假设你工作在一个32位的机器上，你需要将某一个外设寄存器的第X位设置成0(最低位为第0位，最高位为第31位)，将第Y位开始的连续三位设置成110（从高位到低位的顺序），而其他位保持不变。对给定的寄存器值R，及X，Y，编程计算更改后的寄存器值R。

Input

仅一行，包括R,X,Y，以逗号","分隔，R为16进制表示的32位整数，X,Y在0-31之间且Y>=3，(Y-X)的绝对值>=3，保证两次置位不会重合

Output

更改后的寄存器值R（16进制输出）

Sample Input

12345678,0,3

Sample Output

1234567c

Source

Solution:

Implement all the operations by using bit operation, that will be much easier than convert to any other format.

PS, take care of the cin and cout.

Source Code:

[sourcecode language="cpp" collapse="true" padlinenumbers="true"]
//Sat Apr 17 14:43:26 CDT 2010
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

int main( int argc, const char* argv[] )
{
freopen("input.in", "r", stdin);
freopen("output.out", "w", stdout);
int R, X, Y;
char c;
while(cin >> hex >> R >> dec >> c >> X >> c >> Y)
{
R &= ~(1<<(X));
R |= (1<<(Y));
R |= (1<<(Y-1));
R &= ~(1<<(Y-2));

cout << hex << R << endl;
}
fclose(stdin);
fclose(stdout);
return 0;
}
[/sourcecode]