Hdu 5491 The Next【贪心+lowbit】_the next+csdn-优快云博客

本文介绍了一道算法题，任务是找出大于给定整数D的下一个整数Y，该整数的二进制表示中1的数量位于指定范围内。文章提供了解题思路及AC代码。

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The Next

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2184 Accepted Submission(s): 800

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

题目大意：

给你一个数D，再给你两个数S1，S2，设定L表示的是数字D在二进制表示下，1的个数。

问比D大的数Num，使得其满足：S1<=L<=S2的数字是多少。

思路：

①一开始我们可以设定Ans=D+1.那么我们判断Ans的L是多少，如果在【S1，S2】之间的话，直接输出这个Ans即可。否则我们考虑将1的个数进行调整。

②如果存在L>S2,那么我们明显想要去掉1的个数，而且希望尽量去去掉位低的1，而我们去除1的方法只有将Ans增大这一种方法，所以我们考虑找到Ans的最后一个1所在的位子，然后将Ans加上(1<<pos)即可。我们可以用lowbit加速查找。

③如果存在L<S1,那么我们明显想要加1的个数，同时希望增加的1在低位，那么我们从后向前找最低的0的位子，然后将Ans加上(1<<pos)即可。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
#define ll long long int
ll lowbit(ll num)
{
    return num&(-num);
}
int main()
{
    int kase=0;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        ll num;
        int s1,s2;
        scanf("%lld%d%d",&num,&s1,&s2);
        num++;
        while(1)
        {
            int cnt=0;
            ll tmp=num;
            while(tmp)tmp-=lowbit(tmp),cnt++;
            if(cnt>=s1&&cnt<=s2)break;
            else
            {
                if(cnt>s1)
                {
                    num+=lowbit(num);
                }
                else
                {
                    for(int i=0;i<=31;i++)
                    {
                        if(((1<<i)&num)==0)
                        {
                            num+=(1<<i);
                            break;
                        }
                    }
                }
            }
        }
         printf("Case #%d: %lld\n",++kase,num);

    }
}