# PAT 解题报告 1010. Radix: Binary Search以及Overflow检测

## Overview

PAT 解题报告 1010. Radix: 所求的Target有可能是long long 都存不下的数字, AC的方法是使用Binary Search并且需要非常注意Overflow的特殊检测.

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:
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`

## 算法分析：

```//1  if target > s
//0  if target = s
//-1 if target < s int cmp(long long target, const string &s, long long radix) {     long long sv = 0;     long long multip = 1;     for(int i=s.size()-1; i>=0; i--) {
sv += (match(s[i]) * multip); //match(s[i]) return the numerical value of thie character like A = 10
if(sv > target) return -1;// avoid overflow
}
if(sv > target)
return -1;
else if(sv == target)
return 0;
else
return 1;
}
```

## Summary

PAT 解题报告 1010. Radix: 所求的Target有可能是long long 都存不下的数字, AC的方法是使用Binary Search并且需要非常注意Overflow的特殊检测.

Written on April 13, 2013