HDU 5491 The Next

題目連接：http://acm.hdu.edu.cn/showproblem.php?pid=5491

The Next

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1118    Accepted Submission(s): 446

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

 

題目大意是給你一個數，讓你找到一個比它大的數且轉換爲二進制後1的個數在給定的範圍內。

這個題比較簡單，就是對0-1串進行處理，先把給定的數加一，保證結果比給定的數大，而後分兩種狀況，1的個數太少則從低位起遇到第一個0改爲1；1的個數太多則從低位起遇到第一個1加1，而後向前進位。這樣能保證結果是遞增的，當1的個數知足條件是便是最後的結果。

 1 #include <iostream>
 2 #include <cstring>
 3 #include <algorithm>
 4 using namespace std;
 5 
 6 long long ans,t,n,nu,mi,ma,c[50],j,s;
 7 long long fina()
 8 {
 9     long long sum=c[0],te=1;
10     for (int i=1;i<=33;i++)
11     {
12             te*=2;
13             sum=sum+te*c[i];
14     }
15     return sum;
16 }
17 int main()
18 {
19     
20     //cin>>t;
21     scanf("%lld",&t);
22     for (int cas=1;cas<=t;cas++)
23     {
24         //cin>>nu>>mi>>ma;
25         scanf("%lld%lld%lld",&nu,&mi,&ma);
26         nu++;
27         ans=nu;
28         memset(c,0,sizeof(c));
29         j=0,s=0;
30         while (nu!=0)
31         {
32               if (nu%2)
33               s++;
34               c[j++]=nu%2;
35               nu/=2;
36         }
37         j--;
38         while (1)
39         {
40             if (s>=mi&&s<=ma)
41             {
42                   
43                  //cout <<"Case #"<<cas<<": "<<fina()<<endl;
44                  printf("Case #%d: %lld\n",cas,fina());
45                  break;
46             }       
47             if (s<mi)
48             {
49                  for (int i=0;;i++)
50                  {
51                      if (c[i]==0)
52                      {
53                           c[i]=1;
54                           s++;
55                           break;
56                      }
57                  }
58             }
59             else
60             {
61                  int i=0;
62                  while (c[i]==0)
63                  i++;
64                  c[i]++;
65                  while (c[i]==2)
66                  {
67                        c[i]=0;
68                        s--;
69                        c[i+1]++;
70                        i++;
71                  }
72                  s++;
73             }
74         }
75     }
76     return 0;
77 }

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