HUST 1568 Next K-Bit Number（模拟/搜索）

Next K-Bit Number

Time Limit: 1 Sec   Memory Limit: 128 MB
Submissions: 368   Solved: 138

Description

     Our problem is to create a function: uint NextKBitNumber(uint), which given an unsigned integer X with K bits set, returns the immediately larger unsigned integer also with K bits set. For example, if X = 12 (base 10) = 1100(base 2), X has 2 bits set, then  the next number with 2 bits set(NextKBitNumber) is 17(base 10) = 10001(base 2).

Input

      The first line is the number of test cases,there are at most 1001 cases.
      Then each line is a case with an unsigned  number X.
      It is guaranteed that all answers are in unsigned 32-bit Integer.

Output

      For each cases, the first line is “Case #k:”, where k is the case index based 1.
      The second line is  the NextKBitNumber of X.

Sample Input

2
12
7

Sample Output

Case #1:
17
Case #2:
11

题目大意：给定一个32位无符号整数，求出一个数是第一个比它大并且它们转换成二进制时1的个数相等。
思路：有两个，一是模拟，二是暴搜。
模拟就是把这个数换成二进制，然后求出1的个数相等，且比它大的最小的二进制数，再转换成十进制输出。

View Code

 1 # include<stdio.h>
 2 # include<string.h>
 3 # include<iostream>
 4 using namespace std;
 5 int a[50];
 6 int main()
 7 {
 8     int cas,i,j,T,n;
 9     scanf("%d",&T);
10     for(cas=1;cas<=T;cas++){
11         scanf("%d",&n);
12         printf("Case #%d:\n",cas);
13         int cnt=0;
14         int temp;
15         while(n)    //转换成二进制，结果保存在一个数组里边
16         {
17             temp=n%2;
18             a[cnt++]=temp;
19             n=n/2;
20         }
21         int num=0,flag=0; //num表示1的个数，flag表示是否已经找到第一个1
22         for(i=0;i<cnt;i++){
23             if(flag)
24             {
25                 if(a[i]==0){    //求出比原来的二进制大的最小的二进制数
26                     a[i]=1;
27                     for(j=0;j<num-1;j++)
28                         a[j]=1;
29                     for(;j<i;j++)
30                         a[j]=0;
31                     break;
32                 }
33             }
34             if(a[i]) num++;
35             if(!flag && a[i])
36             {
37                 flag=1;
38             }
39         }    
40         if(i>=cnt){    //如果从第一个1开始到最后没有出现过0，这时不得不进位
41             a[cnt++] = 1;
42             for(i=0;i<num-1;i++)
43                 a[i]=1;
44             for(;i<cnt-1;i++)
45                 a[i]=0;
46         }
47         int ans=0;    //转化成十进制
48         for(i=cnt-1;i>=0;i--) ans=ans*2+a[i];
49         printf("%d\n",ans);
50     }
51     return 0;
52 }

暴搜就是从比这个数大的下一个数开始，直到找到一个数的二进制时1的个数和它相等的数，即为结果。

View Code

 1 # include<stdio.h>
 2 # include<string.h>
 3 # include<iostream>
 4 using namespace std;
 5 int num(int x)    //计算整数转化成二进制数时包含1的个数
 6 {
 7     int cnt=0;
 8     while(x)
 9     {
10         if(x%2==1) cnt++;
11         x=x/2;
12     }
13     return cnt;
14 }
15 int main()
16 {
17     int cas,n,i,T;
18     scanf("%d",&T);
19     for(cas=1; cas<=T; cas++)
20     {
21         scanf("%d",&n);
22         int temp = num(n);
23         printf("Case #%d:\n",cas);
24         for(i=n+1;;i++)
25         {
26             if(num(i)==temp) break;
27         }
28         printf("%d\n",i);
29     }
30     return 0;
31 }