AtCoder：A or...or B Problem（思维）

D - A or...or B Problem

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Time limit : 2sec / Memory limit : 256MB

Score : 900 points

Problem Statement

Nukes has an integer that can be represented as the bitwise OR of one or more integers between A and B (inclusive). How many possible candidates of the value of Nukes's integer there are?

Constraints

• 1≤A≤B<260
• A and B are integers.

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Input

The input is given from Standard Input in the following format:

A
B

Output

Print the number of possible candidates of the value of Nukes's integer.

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Sample Input 1

Copy

7
9

Sample Output 1

Copy

4

In this case, A=7 and B=9. There are four integers that can be represented as the bitwise OR of a non-empty subset of {7, 8, 9}: 7, 8, 9 and 15.

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Sample Input 2

Copy

65
98

Sample Output 2

Copy

63

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Sample Input 3

Copy

271828182845904523
314159265358979323

Sample Output 3

Copy

68833183630578410

题意：给定A，B，问[A,B]里取任意个数按位或，结果有多少种。

思路：这题需要找出一个分界点，即找到最高位的B是1，A是0的位置x（最低位从0开始），那么对于所有OR的结果，x处要么是1要么是0，x是0有多少种呢？这里就需要从[A,1<<x)里挑选数进行OR，即有(1<<x)-A种，因为A到(1<<x)-1都可以表示出来；x处为1有几种呢？有两种情况①从(1<<x, B]里挑选，结果就不说了依此类推，②从上述两个区间都挑选，那就是也是(1<<x)-A种，然后上述两种两种情况可能会有重复，需要判断去重，最后判读是否要加上(1<<x)这一种就行了。

# include <bits/stdc++.h>
using namespace std;
 
typedef long long LL;
int main()
{
    LL i, j, a, b, ans=0;
    scanf("%lld%lld",&a,&b);
    if(a == b)
        return 0*puts("1");
    for(i=60; i>=0; --i)
    {
        if((b>>i&1)^(a>>i&1))
            break;
        if(a>>i&1) a ^= 1LL<<i;
    }
    for(j=i-1; j>=0; --j)
        if(b>>j&1)
        {
            ans += (1LL<<j+1)-1;
            break;
        }
    ans += ((1LL<<i)-a)<<1;
    if(j != -1 && a <= (1LL<<j+1)-1) ans -= (1LL<<j+1)-a;
    printf("%lld\n",ans+(i!=0));
    return 0;
}