Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 6789 | Accepted: 4939 |
Description
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
Output
Sample Input
There is no input for this problem
Sample Output
2992 2993 2994 2995 2996 2997 2998 2999 ...
Source
對於一個數x mod n ,可以計算出n進制下x 的末尾數字,再x=x div n;得打除去末尾的數,同理。
#include <stdio.h>
int check(int x,int n)
{
int sum=0;
while(x!=0)
{
sum+=x%n;
x/=n;
}
return sum;
}
int main()
{
int i;
for(i=2992;i<10000;i++)
if ((check(i,10)==check(i,12))&&(check(i,12)==check(i,16))) printf("%d\n",i);
return 0;
}