UVA 10718 Bit Mask

Problem A

Bit Mask

Time Limit

1 Second

In bit-wise expression, mask is a common term. You can get a certain bit-pattern using mask. For example, if you want to make first 4 bits of a 32-bit number zero, you can use 0xFFFFFFF0 as mask and perform a bit-wise AND operation. Here you have to find such a bit-mask.

Consider you are given a 32-bit unsigned integer N. You have to find a mask M such that L ≤ M ≤ U and N OR M is maximum. For example, ifN is 100 and L = 50, U = 60 then M will be 59 and N OR M will be 127 which is maximum. If several value of M satisfies the same criteria then you have to print the minimum value of M.

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

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

Look, a brute force solution may 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

给定一个数n和一个范围l，u在这个范围内找一个最小的m能使n or m最大。

因为涉及到位运算，所以先把n转化为二进制，然后根据或运算的性质来构造m

#include<iostream>using namespace std;int a[64];long long n, l, r, i, cnt;int main(){while (cin >> n >> l >> r){for (i = 0; i < 32; i++, n >>= 1) a[i] = n % 2;for (cnt = 0, i = 31; i >= 0; i--){cnt += (long long) 1 << i;if (!a[i] && cnt <= r || a[i] && cnt <= l) continue;//保证了值最大m最小且在范围内。cnt -= (long long) 1 << i;}cout << cnt << endl;}return 0;}