Specialized Four-Digit Numbers
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4978 Accepted Submission(s): 3597
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.)
#include<cstdio>
#include<iostream>
#include<cstring>
const int maxn=1000;
using namespace std;
int t[maxn],A[maxn];
char str1[maxn],str2[maxn];
int n,m;
void solve(char *str1,char *str2,int n,int m)
{
int i,len,k;
len=strlen(str1);
for(i=len;i>=0;i--)
{
t[len-1-i]=str1[i]-(str1[i]<58?48:str1[i]<97?55:61);
//小於58說明是48-57是數字
//小於97是65-90爲大寫英文字母
//大於97是97-122爲小寫英文字母!
//同時也說明了0-9存的是數字
//10-25存的是大寫字母
//26-61存的是小寫字母
}
for(k=0;len;)
{
for(i=len;i>=1;i--)
{
t[i-1]+=t[i]%m*n;
t[i]/=m;
}
A[k++]=t[0]%m;
t[0]/=m;
while(len>0&&!t[len-1])
{
len--;
}
}
str2[k]=NULL;
for(i=0;i<k;i++)
{
str2[k-1-i]=A[i]+(A[i]<10?48:A[i]<36?55:61);
}
}
int main()
{
int i,j,sum1,sum2,sum3;
for(i=1000;i<=9999;i++)
{
sum1=sum2=sum3=0;
sprintf(str1,"%d",i);
for(j=0;j<strlen(str1);j++)
{
sum1+=str1[j]-'0';
}
solve(str1,str2,10,16);
for(j=0;j<strlen(str2);j++)
{
if(str2[j]>='A'&&str2[j]<='Z')
{
sum2+=str2[j]-'A'+10;
}
else
{
sum2+=str2[j]-'0';
}
}
solve(str1,str2,10,12);
for(j=0;j<strlen(str2);j++)
{
if(str2[j]>='A'&&str2[j]<='Z')
{
sum3+=str2[j]-'A'+10;
}
else
{
sum3+=str2[j]-'0';
}
}
if(sum1==sum2&&sum1==sum3)
{
cout<<str1<<endl;
}
}
}