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  • poj2196

    Specialized Four-Digit Numbers
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 7238   Accepted: 5285

    Description

    Find and list all four-digit numbers in decimal notation that have the property that the sum of its four digits equals the sum of its digits when represented in hexadecimal (base 16) notation and also equals the sum of its digits when represented in duodecimal (base 12) notation.
    For example, the number 2991 has the sum of (decimal) digits 2+9+9+1 = 21. Since 2991 = 1*1728 + 8*144 + 9*12 + 3, its duodecimal representation is 189312, and these digits also sum up to 21. But in hexadecimal 2991 is BAF16, and 11+10+15 = 36, so 2991 should be rejected by your program.
    The next number (2992), however, has digits that sum to 22 in all three representations (including BB016), so 2992 should be on the listed output. (We don't want decimal numbers with fewer than four digits -- excluding leading zeroes -- so that 2992 is the first correct answer.)

    Input

    There is no input for this problem

    Output

    Your output is to be 2992 and all larger four-digit numbers that satisfy the requirements (in strictly increasing order), each on a separate line with no leading or trailing blanks, ending with a new-line character. There are to be no blank lines in the output. The first few lines of the output are shown below.

    Sample Input

    There is no input for this problem

    Sample Output

    2992
    2993
    2994
    2995
    2996
    2997
    2998
    2999
    ...

    Source

    Pacific Northwest 2004
     1 #include <iostream>
     2 
     3 using namespace std;
     4 
     5 
     6 int digits(int x)
     7 {
     8     int a = x/1000;
     9     int sum = 0;
    10     sum +=a;
    11     a = x%1000/100;
    12     sum +=a;
    13     a = x%1000%100/10;
    14     sum +=a;
    15     a = x%1000%100%10;
    16     sum +=a;
    17     return sum;
    18 }
    19 int digits_12(int x)
    20 {
    21     int num_12[4];
    22     for(int i =0 ;i<4;i++)
    23     {
    24         num_12[i]=x%12;
    25         x=x/12;
    26     }
    27     return num_12[0]+num_12[1]+num_12[2]+num_12[3];
    28 }
    29 int digits_16(int x)
    30 {
    31     int num_16[4];
    32     for(int i =0 ;i<4;i++)
    33     {
    34         num_16[i]=x%16;
    35         x=x/16;
    36     }
    37     return num_16[0]+num_16[1]+num_16[2]+num_16[3];
    38 }
    39 int main()
    40 {
    41 
    42     int num_16[4];
    43     int x = 2992;
    44     for(int i=x;i<=9999;i++)
    45     {
    46         int sum = digits(i);
    47         int sum_12 = digits_12(i);
    48         int sum_16 = digits_16(i);
    49         if(sum ==sum_12&&sum==sum_16)
    50         {
    51             cout<<i<<endl;
    52         }
    53     }
    54     return 0;
    55 }
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  • 原文地址:https://www.cnblogs.com/jhldreams/p/3751889.html
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