hdu 5491 The Next（暴力枚举）

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

 

Source

2015 ACM/ICPC Asia Regional Hefei Online

 

不想写，找了个别人的.

思路：

考虑通过比较当前的1的数目来进行最小的数的修正。

先将D加1，然后计算出D的1的数目tot，这时候比较tot和s1,s2的大小，这时候有两种情况：

1、tot<s1，这时我需要增加1的数目，因为要最小，所以我从右开始找到一个0（位置假设为i），将D加上2^i。

2、tot>s2，这时我需要减少1的数目，所以我从右开始找到一个1，将D加上2^i。

如此循环上述过程，直到符合tot符合s1,s2的条件。

上面操作的原因是：我加上2^i，就是将那一位由1变为0或者由0变为1，而我是从右边开始寻找的，可以保证每次所做的改变都是最小的。同时我的D是一直增加的，所以当条件满足时，就是那个最小的。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<bitset>
#include<map>
#include<vector>
#include<stdlib.h>
#include <stack>
using namespace std;
int dirx[]={0,0,-1,1};
int diry[]={-1,1,0,0};
#define PI acos(-1.0)
#define max(a,b) (a) > (b) ? (a) : (b)
#define min(a,b) (a) < (b) ? (a) : (b)
#define ll long long
#define eps 1e-10
#define MOD 1000000007
#define N 100
#define inf 1e12
ll d,s1,s2;
ll num[N];
ll k;
ll count_(ll dd){
   memset(num,0,sizeof(num));
   ll cnt=0;
   k=0;
       while(dd){
           if(dd&1) cnt++;
           num[k++]=(dd&1);
           dd>>=1;
       }
       num[k++]=0;

   return cnt;
}
int main()
{
   ll ac=0;
   ll t;
   scanf("%I64d",&t);
   while(t--){
       scanf("%I64d%I64d%I64d",&d,&s1,&s2);

       ll dd=d+1;

       while(1){
           ll tmp=count_(dd);

           if(tmp<s1){
              for(ll i=0;i<k;i++){
                  if(num[i]==0){
                     dd+=(1<<i);
                     break;
                  }
              }
           }else if(tmp>s2){
              for(ll i=0;i<k;i++){
                  if(num[i]==1){
                     dd+=(1<<i);
                     break;
                  }
              }
           }else{
              break;
           }
       }
       printf("Case #%I64d: ",++ac);
       printf("%I64d
",dd);

   }
    return 0;
}