UVa 10718 bit Mask: greedy & bit operations

10718-bit Mask

Time limit:3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem= 1659

In bit-wise expression, mask is a common term. You can get a certain bit-pattern using mask. For example, if your want to make the a 32-bit number zero, can use 0XFFFFFFF0 as mask and perform a bit-wi SE and operation. Here's have to find such a bit-mask.

Consider you are given a 32-bit unsigned integern. You are have to find a maskm such which L≤m≤u and N OR M is maximum. For example, IfN is Andl = x, U = Thenm'll be to be ANDN 127 OR M'll be which. If several value OfM satisfies the same criteria then you have to print the minimum value OfM.

Input
Each input starts with 3 unsigned integers N, L, and U where l≤u. Input is terminated by EOF.

Output
For each input, print in a line the minimum value of M, which makesn OR m maximum.

Look, a brute force solution could not end within the time limit.

Sample Input	Output for Sample Input
100 50 60 100 50 50 100 0 100 1 0 100 15 1 15	59 50 27 100 1

Greedy idea:

From high to low to consider,

If the first bit of n is 0, then M needs to be as much as 1 in this position, and after this bit becomes 1 m<=u;
If the first bit of n is 1, then M is required to be 0 in this position, but m cannot be too small to be m<l when L is in this position for 1 o'clock.

Complete code:

/*0.013s*/
     
#include <cstdio>  
typedef unsigned int UI;  
     
int main ()  
{  
    ui n, L, u, M, temp;  
    int i;  
    while (~SCANF ("%u%u%u", &n, &l, &u))  
    {  
        m = 0;  
        for (i =; I >= 0-i)  
        {  
            ///if n is 0, then M should try to be 1 in this bit, and after this one is 1 m<=u  
            ///if n is 1, then M should try to be 0 in this position, But m cannot be too small as L in this one bit for 1 o'clock m<l  
            ///Notice n<l this situation  
            temp = m | (1u << i); M + (1u << i) of///bitwise operation ((  
            n >> i) & 1) = = 0 && temp <= u | | (l >> i) & 1 && m < l) m = temp;  
        }  
        printf ("%u\n", m);  
    }  
    return 0;  
}

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