Binary Numbers

Binary Numbers

时间: 1ms        内存:64M

描述:

Given a positive integer n, find the positions of all 1's in its binary representation. The position of the least significant bit is 0. Example The positions of 1's in the binary representation of 13 are 0, 2, 3. Task Write a program which for each data set: * reads a positive integer n, * computes the positions of 1's in the binary representation of n, * writes the result.

输入:

The first line of the input contains exactly one positive integer d equal to the number of data sets, 1 ≤ d ≤ 10. The data sets follow. Each data set consists of exactly one line containing exactly one integer n, 1 ≤ n ≤ 106.

输出:

The output should consists of exactly d lines, one line for each data set. Line i, 1 ≤ i ≤ d, should contain increasing sequence of integers separated by single spaces - the positions of 1's in the binary representation of the i-th input number. Do not output any spaces in the end of a line.

示例输入:

1
13

示例输出:

0 2 3

提示:

参考答案(内存最优[1032]):

#include <stdio.h>
int main()
{
          int a[10],i,j,k=0,p=0,n;
          scanf("%d",&n);
          for(i=0;i<n;i++)
                scanf("%d",&a[i]);
          for(i=0;i<n;i++)
         {
                k=0;
                p=0;
                for(j=0;j<sizeof(int)*8;j++)
               {
                         if(a[i]%2!=0){ 
                                if(p==0)
                                           printf("%d",k);
                                else
                                           printf(" %d",k);
                                p++;
                         }
                        a[i]=a[i]>>1;
                        k++;
               }
               printf("\n");
         }
         return 0;
}

参考答案(时间最优[0]):

#include"stdio.h"
int a[100000]={0};
int main()
{
    int t,i,j,n,k;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=0; n!=1; i++)
        {
            a[i]=n%2;
            n/=2;
        }
        for(j=0; j<i; j++)
            if(a[j]==1)
                printf("%d ",j);
        printf("%d\n",i);
    }
    return 0;
}

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