uva 10718 Bit Mask （位运算）
|
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, 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 |
59 |
Member of Elite Problemsetters' Panel
| N的值最大，M的该位应考虑置为1，然后判断M的该位为1时的可能取值区间[lmin, lmax]，如果区间[lmin, lmax]与区间[L, U]有交集，则说明该位可置为1；如果N的该位为1，为使M的值尽可能小，M的该位应考虑置为0，然后判断可能取值区间与[L, U]是否有交集，如果没有交集，该位置为1。
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
long long n,l,r,a[100],b[100];
int cnta,cntb;
void get(){
long long x=n;
cnta=0;cntb=0;
while(x>0){
a[cnta++]=x%2;
x/=2;
}
x=r;
while(x>0){
b[cntb++]=x%2;
x/=2;
}
}
void computing(){
long long ans=0;
if(cntb>cnta){
for(int i=cntb-1;i>cnta-1;i--){
long long tmp=(long long)1<<(long long)i;//此处不强转，会超出int范围，wa了几次
if(b[i]==1){
l-=tmp;
r-=tmp;
ans+=tmp;
}
}
}
for(int i=cnta-1;i>=0;i--){
long long tmp=(long long)1<<(long long)i;
if(tmp<=l){
l-=tmp;
r-=tmp;
ans+=tmp;
}
else if(a[i]==0 && tmp<=r){
l-=tmp;
r-=tmp;
ans+=tmp;
}
}
printf("%lld\n",ans);
}
int main(){
while(scanf("%lld%lld%lld",&n,&l,&r)!=EOF){
get();
computing();
}
return 0;
}
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