Given a pair of positive integers, for example, 6 and 110, can this equation 6 = 110 be true? The answer is "yes", if 6 is a decimal number and 110 is a binary number.
Now for any pair of positive integers N1 and N2, your task is to find the radix of one number while that of the other is given.
Input Specification:
Each input file contains one test case. Each case occupies a line which contains 4 positive integers:
N1 N2 tag radix
Here N1 and N2 each has no more than 10 digits. A digit is less than
its radix and is chosen from the set {0-9, a-z} where 0-9 represent the
decimal numbers 0-9, and a-z represent the decimal numbers 10-35. The
last number "radix" is the radix of N1 if "tag" is 1, or of N2 if "tag"
is 2.
Output Specification:
For each test case, print in one line the radix of the other number
so that the equation N1 = N2 is true. If the equation is impossible,
print "Impossible". If the solution is not unique, output the smallest
possible radix.
Sample Input 1:
6 110 1 10
Sample Output 1:
2
Sample Input 2:
1 ab 1 2
Sample Output 2:
Impossible 现在是只能拿18分，感觉自己题意理解的有问题，过几天再看看吧，今天实在是看不下去了....
#include <iostream>
#include <cmath>
#include <cstring> using namespace std; long long int translate(string a,int r){
long long int a10=;
int x;
int n=a.length();
int c=;
for(int i=n-;i>=;i--){
if(a[i]>)
x=a[i]-;
else{
x=a[i]-;
}
a10=a10+x*pow(r,c);
c++;
}
return a10;
} int main()
{
string a,b;
long long int a10=,b10=;
int n,r;
int flag=;
cin>>a>>b>>n>>r;
if(n==){
a10=translate(a,r);
int i;
if(b.length()==) i=translate(b,)+;
else i=;
for(;i<;i++){
if(translate(b,i)==a10){
cout<<i<<endl;
flag=;
break;
} }
if(flag==)
cout<<"Impossible"<<endl; }
else if(n==){
b10=translate(b,r);
int i;
if(a.length()==) i=translate(a,)+;
else i=;
for(;i<;i++){
if(translate(a,i)==b10){
cout<<i<<endl;
flag=;
break;
} }
if(flag==)
cout<<"Impossible"<<endl;
}
return ;
}