Pat Class A exercises-Radix (25)-Two points find

1010. Radix (+) time limit MS
Memory Limit 65536 KB
Code length limit 16000 B
Program Standard author CHEN, Yue

Given a pair of positive integers, for example, 6 and +, can this equation 6 = + be true? The answer is ' yes ', if 6 is a decimal number and a binary number.

Now for any pair of positive integers N1 and N2, your task was to find the radix of one Iven.

Input Specification:

Each input file contains the one test case. Occupies a line which contains 4 positive integers:
N1 N2 Tag Radix
Here N1 and N2 each have no more than digits. A Digit is less than it 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, the equation N1 = N2 is true. If the equation is impossible, print "impossible". If The solution is not unique, the output of 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

/** Test Instructions: N1 n2 can be equal in some kind of binary, to find out that the binary tag = 1 radix for the N1 = 2 for the N2 of the input: the smallest of the system is definitely the largest digit +1 in N1, the largest system is not 26, but N1+1, said N2 with one can Represents N1 so it is not possible to use a For loop enumeration, but to use a binary lookup to note the range of data to use long long note that the N2 may overflow under X, the lookup area should be fixed to the left half */#include <bits/stdc++.h> using
namespace Std;
typedef pair<int,int> P;
typedef long Long LL;
const int MAXN = 2000+10;

const int INF = 0X3F3F3F3F;
    ll Change (char c) {if (c >= ' 0 ' && C <= ' 9 ') return C ' 0 ';
else return C ' a ' +10;
    } ll Getval (string S,int x) {///the value converted to decimal in the X-binary is ll res = 0;
        for (int i = 0;i<s.length (); ++i) {res *= x;
        Res + = Change (S[i]);

if (Res < 0) return-1;///here Note that res may overflow} return res;
    } int Main () {string n1,n2,t;
    int Tag,radix;
    Cin>> n1>>n2>>tag>>radix;
        if (tag = = 2) {///Let N1 always indicate the reference t = N1;
        N1 = n2;
    N2 = t;
    } ll N1val = Getval (N1,radix);
    ll low = 0, high, mid; for (int i = 0;i<n2.length (); ++i) {if(Change (N2[i]) > Low) low = Change (N2[i]);
    } low++;

    High = n1val+1;
    bool F = 0;
        while (low <= high) {mid = (Low+high)/2;
        ll midval = Getval (N2,mid);
        if (Midval < 0 | | midval > N1val) high = mid-1;
        else if (Midval < n1val) Low = mid +1;
            else if (Midval = N1val) {f = 1;
        Break
    }} if (f) cout << mid;

    else cout<< "impossible";

return 0;
 }