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  • HDU 1390 Binary Numbers

    题目地址:

    http://acm.hdu.edu.cn/showproblem.php?pid=1390

    题目描述:

     Binary Numbers

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 2834    Accepted Submission(s): 1758

    Problem Description

    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.
     

    Input

    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 <= 10^6.
     

    Output

    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.
     
    Sample Input
    1 13
     

    Sample Output

    0 2 3

     题意:

    输出2进制数位为1的位置,从低到高。

    题解:

    这题主要注意两点:
    1、二进制的低位高位问题(与10进制是de一样),11110000;1是高位,0是低位
    2、注意格式问题,第一个余数为1的输出 "%d" ,其余后序余数为1的输出 " %d",空格不对会出Presentation Error错误  
     
     1 #include <iostream>
     2 using namespace std;
     3 
     4 int main(void)
     5 {
     6     int n,j;
     7     int num;
     8 
     9     cin>>n;
    10     while(n--)
    11     {
    12         j=0;
    13         cin>>num;
    14         while(num>0)
    15         {
    16             if(num%2==1&&num/2==0)
    17             {
    18                 cout<<j;
    19             }else if(num%2==1&&num/2>0)
    20             {
    21                 cout<<j<<" ";
    22             }
    23 
    24             num/=2;
    25             j++;
    26         }
    27         cout<<endl;
    28     }
    29     return 0;
    30 }
    View Code
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  • 原文地址:https://www.cnblogs.com/samjustin/p/4567331.html
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