题目链接：

The Next

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1766    Accepted Submission(s): 669

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

题意：

给一个D,现在让求最小的ans>D,且ans的二进制中1的个数是[s1,s2];

思路：

套路题,从高位到低位按位贪心,对于当前位,如果为1,那么这一位就一定要取1,然后d,s1,s2都变小了;

如果当前位为0,那么就要看取最多个数的1能得到的数如果大于当前的数,那么还可以取0;

否则就要取1了;就这样贪心,具体的可以看代码;

AC代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn=1e5+10;
int l,r;
LL d,ans,dp[40];
void dfs(LL cur,int hi,int lo,int pos)
{
    if(hi<=0||pos<0)return ;
    if(cur>=dp[pos]){ans+=dp[pos],dfs(cur-dp[pos],hi-1,lo-1,pos-1);}
    else
    {
        if(lo==pos+1)
        {
            ans+=dp[lo]-1;
            return ;
        }
        else
        {
            LL num=0;
            for(int i=pos-1;i>=max(0,pos-hi);i--)num+=dp[i];
            if(num>cur)dfs(cur,hi,lo,pos-1);
            else
            {
                ans+=dp[pos];
                lo--;
                if(lo>0)ans+=dp[lo]-1;
                return ;
            }
        }
    }
}
int main()
{
    int t,Case=0;
    scanf("%d",&t);
    dp[0]=1;
    for(int i=1;i<=33;i++)dp[i]=dp[i-1]*2;
    while(t--)
    {
        scanf("%lld%d%d",&d,&l,&r);
        ans=0;
        dfs(d,r,l,32);
        if(d==0)
        {
            if(l==0)ans=1;
            else ans=dp[l]-1;
        }
        printf("Case #%d: %lld\n",++Case,ans);
    }
    return 0;
}