Bits - CodeForces 485C

最新推荐文章于 2020-07-31 16:48:30 发布

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最新推荐文章于 2020-07-31 16:48:30 发布

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本文详细解析如何使用C++解决Codeforces平台上的力赛题，题目要求在给定区间内找到二进制下1的个数最多的整数，且在数量相同时选择最小的数。通过预设数组存储特定二进制数并累加，直至超过上限，进而找到答案。

Let's denote as $\text{[math]}$ the number of bits set ('1' bits) in the binary representation of the non-negative integer x.

You are given multiple queries consisting of pairs of integers l and r. For each query, find the x, such that l ≤ x ≤ r, and $\text{[math]}$ 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 integers li, ri — the arguments for the corresponding query (0 ≤ li ≤ ri ≤ 1018).

Output

For each query print the answer in a separate line.

   Sample test(s)
 

     input
   
3
1 2
2 4
1 10

     output
   
1
3
7

Note

The binary representations of numbers from 1 to 10 are listed below:

110 = 12

210 = 102

310 = 112

410 = 1002

510 = 1012

610 = 1102

710 = 1112

810 = 10002

910 = 10012

1010 = 10102

题目链接：http://codeforces.com/problemset/problem/485/C
题目大意：给定一个区间[l, r]，求这个区间里的数在转化为二进制的情况下，1的个数最多的数字；同时，当1的个数相同的时候，输出数最小的那个。
题目思路：先设置一个数组，用来存储1、10、100、1000这些二进制数，然后对这些数进行累加，直到大于等于r为止。此时的sum就是1的个数最多的那个数，但是有可能不在这个区间中，所以我们需要进行判断。假如他大于r的话，就减去最后相加的那个数，这样可以保证得出的结果是区间中1最多的但是数值最小的。

C++源代码如下：

#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <iomanip>
#include <cstring>
#include <iostream>
#include <algorithm>
#define For(i, n) for (int i = 1; i <= n; i++)
#define ForK(i, k, n) for (int i = k; i <= n; i++)
#define ForD(i, n) for (int i = n; i >= 0; i--)
#define Rep(i, n) for (int i = 0; i < n; i++)
#define RepD(i, n) for (int i = n; i >= 0; i--)
#define MemI(a) memset(a, 0, sizeof(a))
#define MemC(a) memset(a, '\0', sizeof(a))
#define PI acos(-1)
#define eps 1e-8
#define inf 0x3f3f3f3f
typedef long long ll;
using namespace std;

ll bit[65];
int main()
{
    bit[0] = 1;
    for (int i = 1; i <= 63; i++)
    {
        bit[i] = 2 * bit[i - 1];
    }
    int t;
    ll l, r, sum;
    scanf("%d", &t);
    while (t--)
    {
        sum = 0;
        scanf("%I64d%I64d", &l, &r);
        int i;
        for (i = 0; sum <= r; i++)
        {
            sum += bit[i];
        }
        i--;
        while (sum > r)
        {
            sum -= bit[i];
            if (sum < l)
                sum += bit[i];
            i--;
        }
        printf("%I64d\n", sum);
    }
    return 0;
}