HDU 1197 Specialized Four-Digit Numbers(进制转换+暴力)

Specialized Four-Digit Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5621    Accepted Submission(s): 4063

Problem 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 1893(12), 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

29922993299429952996299729982999

 

题解：将给定的数字转化为10进制 12进制 16进制,并求出转化进制后每一位的和,若相等的话就输出

AC代码：

#include<iostream>#include<cstdlib>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<cstdlib>#include<algorithm>typedef long long LL;using namespace std;//将给定的数字转化为10进制 12进制 16进制,并求出转化进制后每一位的和,若相等的话就输出int check(int i,int n){int s=0;while(i>0){s=s+i%n;i=i/n;}return s;}int main(){int i=2992;int n;for(i=2992;i<=9999;i++){if(check(i,10)==check(i,12)&&check(i,10)==check(i,16))printf("%d\n",i);}   return 0;}