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                            uva 10718 Bit Mask

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, if N 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.

题目大意：每组数据包含三个有效数据：1）N； 2）L ； 3）U   要在L～U之间找出M和N做或运算得出的值最大的，并且本身值最小的数。

解题思路：一开始，直接用或运算+暴力，毫不犹豫的超时了。然后用了另一种方法，先将N转化为二进制存在一个数组中，然后从高位开始往下遍历，构造M，若当前位为0，则在不超过上限的情况下，M的该位为1；若当前位为1，则在达不到下限的情况下，将M的该位赋为1。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#define M 32
typedef long long ll;
using namespace std;
ll N, L, U, B[M + 1];
int A[M + 1];
int change(int *T, ll a) {
	for (int i = 0; i < M; i++) {
		T[i] = a % 2;
		a /= 2;
	}
}
int main() {
	B[0] = 1;
	for (int i = 1; i < M; i++) {
		B[i] = B[i - 1] * 2;
	}
	while (scanf("%lld %lld %lld", &N, &L, &U) == 3) {
		ll ans = 0, temp;
		change(A, N);
		for (int i = M - 1; i >= 0; i--) {
			if (!A[i]) {
				temp = ans + B[i];
				if (temp <= U) { //当N的该位为0，则在不超过上限的情况下，选1
					ans += B[i];
				}
			}
			else {
				temp = ans + B[i] - 1;
				if (temp < L) { //如果该位选1，达不到下限，则选1，若可以达到下限，为了保持ans最小，就选0
					ans += B[i];
				}
			}
		}
		printf("%lld\n", ans);
	}
	return 0;
}

uva 10718 Bit Mask （位运算）

标签：acm   c语言   uva   

原文地址：http://blog.csdn.net/llx523113241/article/details/44132279