hdoj 5491 The Next 【lowbit 的使用】

最新推荐文章于 2016-01-13 10:29:13 发布

原创最新推荐文章于 2016-01-13 10:29:13 发布 · 594 阅读

0 ·

CC 4.0 BY-SA版权

二进制和位运算专栏收录该内容

6 篇文章

订阅专栏

本文介绍了一种算法，用于找到给定范围内下一个满足特定条件的WYH数。

The Next

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 926 Accepted Submission(s): 381

Problem Description

Let

L denote the number of 1s in integer

D’s binary representation. Given two integers

S1 and

S2, we call

D a WYH number if

S1≤L≤S2.
With a given

D, we would like to find the next WYH number

Y, which is JUST larger than

D. In other words,

Y is the smallest WYH number among the numbers larger than

D. Please write a program to solve this problem.

Input

The first line of input contains a number

T indicating the number of test cases (

T≤300000).
Each test case consists of three integers

S1, and

S2, as described above. It is guaranteed that

0≤D<231 and

D is a WYH number.

Output

For each test case, output a single line consisting of “Case #X: Y”.

X is the test case number starting from 1.

Y is the next WYH number.

Sample Input

Sample Output

Case #1: 12
Case #2: 25
Case #3: 17

WYH数的定义：若x的二进制中1的个数num满足s1 <= num <= s2 则x是一个WYH数。

题意：给一个WYH数D、s1和s2，让你求出第一个大于D的WYH数。题目保证存在解。

lowbit性质：

对于满足[ x, x + lowbit(x)) 的数，二进制中1的个数随数的增大是单调增的（注意区间不包括x + lowbit(x) 而且单调增不是严格的）。

思路：每次判断区间[x ~ x+lowbit(x))里面是否存在满足条件的WYH ，如果存在则在该区间找最终结果即叠加上最少需要增加的低位1，反之找下一个区间即[x+lowbit(x), x + lowbit(x) + lowbit(x+lowbit(x)) )。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
using namespace std;
LL lowbit(LL x){
    return x & -x;
}
int Count(LL x)//1的个数
{
    int sum = 0;
    while(x)
    {
        if(x & 1)
            sum++;
        x >>= 1;
    }
    return sum;
}
int main()
{
    int t, k = 1;
    LL d;
    int s1, s2;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%lld%d%d", &d, &s1, &s2);
        LL  ans = d + 1;
        int cnt = Count(ans);
        LL lb = lowbit(ans);
        while(s1 > cnt || s2 < cnt)
        {
            int addnum = Count(lb-1);
            if(s1 > cnt + addnum || cnt > s2)
            {
                ans += lb;//继续下一个区间
                cnt = Count(ans);
                lb = lowbit(ans);
            }
            else
            {
                ans += (1<<(max(cnt, s1)-cnt))-1;//叠加低位1
                break;
            }
        }
        printf("Case #%d: %lld\n", k++, ans);
    }
    return 0;
}