标签：des style c class blog code
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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标签：des style c class blog code
原文地址：http://www.cnblogs.com/jhldreams/p/3751889.html
