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.
|
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 |
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;
}