1014. Specialized Four-Dig
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1014. Specialized Four-Dig
Constraints
Time Limit: 1 secs, Memory Limit: 32 MB
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
进制转化加简单计算问题:十进制转向十二和十六进制。
// Problem#: 1014
// Submission#: 1950974
// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License
// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/
// All Copyright reserved by Informatic Lab of Sun Yat-sen University
#include<stdio.h>
int judge1(int i,int result);
int judge2(int i,int result);
int main(){
int flag=0;
int i;
int result;
for(i=2992;i<10000;i++){
int temp=i;
result=0;
while(temp>0){
result+=temp%10;
temp/=10;
}
flag+=judge1(i,result);
flag+=judge2(i,result);
if(flag==2){
printf("%d\n",i);
}
flag=0;
}
return 0;
}
int judge1(int i,int result){
int temp=0;
while(i>0){
temp+=i%16;
i/=16;
}
if(temp==result){
return 1;
}
else{
return 0;
}
}
int judge2(int i,int result){
int temp=0;
while(i>0){
temp+=i%12;
i/=12;
}
if(temp==result){
return 1;
}
else{
return 0;
}
}
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