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 3 11 2 4 22 3 3 15 2 5   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;
}