hdoj 5491 The Next 【lowbit 的使用】

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 D, 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

311 2 422 3 315 2 5

 

Sample Output

Case #1: 12Case #2: 25Case #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 longusing 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;}