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。

# include<stdio.h>
# include<string.h>
# include<iostream>
using namespace std;
int a[50];
int main()
{
    int cas,i,j,T,n;
    scanf("%d",&T);
    for(cas=1;cas<=T;cas++){
        scanf("%d",&n);
        printf("Case #%d:\n",cas);
        int cnt=0;
        int temp;
        while(n)    //转换成二进制，结果保存在一个数组里边
        {
            temp=n%2;
            a[cnt++]=temp;
            n=n/2;
        }
        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
        for(i=0;i<cnt;i++){
            if(flag)
            {
                if(a[i]==0){    //求出比原来的二进制大的最小的二进制数
                    a[i]=1;
                    for(j=0;j<num-1;j++)
                        a[j]=1;
                    for(;j<i;j++)
                        a[j]=0;
                    break;
                }
            }
            if(a[i]) num++;
            if(!flag && a[i])
            {
                flag=1;
            }
        }    
        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
            a[cnt++] = 1;
            for(i=0;i<num-1;i++)
                a[i]=1;
            for(;i<cnt-1;i++)
                a[i]=0;
        }
        int ans=0;    //转化成十进制
        for(i=cnt-1;i>=0;i--) ans=ans*2+a[i];
        printf("%d\n",ans);
    }
    return 0;
}

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。

# include<stdio.h>
# include<string.h>
# include<iostream>
using namespace std;
int num(int x)    //计算整数转化成二进制数时包含1的个数
{
    int cnt=0;
    while(x)
    {
        if(x%2==1) cnt++;
        x=x/2;
    }
    return cnt;
}
int main()
{
    int cas,n,i,T;
    scanf("%d",&T);
    for(cas=1; cas<=T; cas++)
    {
        scanf("%d",&n);
        int temp = num(n);
        printf("Case #%d:\n",cas);
        for(i=n+1;;i++)
        {
            if(num(i)==temp) break;
        }
        printf("%d\n",i);
    }
    return 0;
}