Codeforces Round #276 (Div. 1) A. Bits

Codeforces Round #276 A

Let's denote as the number of bits set ('1' bits) in the binary representation of the non-negative integerx.

You are given multiple queries consisting of pairs of integers l and r. For each query, find thex, such thatl ≤ x ≤ r, and is maximum possible. If there are multiple such numbers find the smallest of them.

Input

The first line contains integer n — the number of queries (1 ≤ n ≤ 10000).

Each of the following n lines contain two integersl_i, r_i — the arguments for the corresponding query (0 ≤ l_i ≤ r_i ≤ 10¹⁸).

Output

For each query print the answer in a separate line.

题目大意：

在范围[a,b]中寻找二进制位1最多的最小数

解题思路：

   假设开始二进制每一位都为1，从高位i开始枚举，如果当前数大于b，且减去1<<i后仍大于等于a，就减1<<i，直到当前数在[a,b]之间时，输出当前数（由于是从高位开始减，所以当前数是最小的）

 

代码：

#include<iostream>
#include<map>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<cstdio>
using namespace std;
typedef long long LL;;

LL bit[65];

int main() {
    bit[0] = 1;
    for(int i = 1; i <= 60; i++) bit[i] = bit[i-1] << 1; 
    int n;
    scanf("%d", &n);
    LL a, b, ans;
    while(n--) {
        scanf("%lld %lld", &a, &b);
        ans = (1LL << 61) - 1;
        for(int i = 60; i >= 0; i--) {
            if(ans > b && ans - bit[i] >= a) ans -= bit[i];
            if(ans >= a && ans <= b) break;
        }
        cout << ans << endl;
    }
    return 0;
}