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, 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

 

很容易想到要用貪心，但是真的貪起來還挺麻煩的，n or m要最大， m在[L,U]內，m要儘可能小，

先考慮n>m的情況，要or運算最大，自然把n轉化爲二進制後，出現零的位置，儘量由m轉化爲二進制來填補；

所以可以從最高位開始，遇到n的二進制位爲0的，判斷m在該位爲1是否會大於U，但是最後雖然可以使n or m最大，m卻可能小於L，這樣貪心只保證了m<=U；

因爲我們忽略了m的二進制位爲0的情況；

有個常識需要注意A=1000....（n個0）B=11111（n個1） A=B+1 A>B ;所以爲了避免m小於L的情況出現，我們只需要判斷，m二進制位後面全取1是否小於L，是的話，當前位必須取1；

最後整理一下思路就是，初始化m的所有二進制位爲0，我們從高位到低位找到哪些些位置儘量爲1（n or m 大），那些位置位置必須爲1（m>=L),

#include <stdio.h>
#include <string.h>
int main()
{ 
 long long ans,i,j,m,n,l,u,a[100],b[100],exp[100],size=32;
 exp[1]=1;
 for (i=2;i<=size;i++)
 exp[i]=exp[i-1]*2;
 while (scanf("%lld%lld%lld",&n,&l,&u)!=EOF)
 {  
  memset(a,0,sizeof(a));
  memset(b,0,sizeof(b));
  i=0;
  while (n)
  {
   a[++i]=n%2; 
   n=n/2;
  }

  for (i=size;i>=1;i--)
  {
   if (a[i]==0)
   { 
    b[i]=1;
    m=0;
    for (j=1;j<=size;j++)
    m=m+b[j]*exp[j];
    if (m>u) b[i]=0;
   }
   if (b[i]==0)
   {
    m=0;
	for (j=32;j>i;j--)
    m=m+b[j]*exp[j];
	if (m+exp[i]-1<l) b[i]=1; 
   }
  }
 ans=0;
 for (i=1;i<=32;i++)
 ans=ans+b[i]*exp[i];
 printf("%lld\n",ans);
 }
 return 0;
}