PAT 1010. Radix (25)

PAT原题在此

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

• 这道题考察的点包括
  
  • 字符串与数的转换
  • 进制转换
  • 二分查找

细节准备

因为这里出的数极有可能不是十进制的数，所以需要通过string类型存储，为之后转换成数做准备；同时这道题进制上限太高，出现zzzzz这种数极有可能溢出，用 long long 比较好

字符串与数的转换

int toNum(char& a){    if (a >= '0' && a <= '9'){        return a - '0';    }    else{        return a - 'a' + 10;    }}

进制转换

typedef long long lint;//将某个数转换成十进制数，使得同进制下进行直接比较int toDecimal(string& num, int& base){    lint sum = 0;    for (auto& x : num){        sum *= base;        sum += toNum(x);    }    return sum;}

二分查找

//假设给出进制的数是a，那么得到a的值后，我们搜索合适的进制使得a==b    lint low, high=a+1;//a+1就够了，只要b>0    while (low <= high){//二分查找形式多样，一定要注意边界        lint mid = (low + high) / 2;        lint res = toDecimal(b, mid);        if (res == aa){            mark = true;//标记找到            bbase = mid;//记录进制            break;        }        else if (res > aa){            low = mid + 1;        }        else{            high = mid - 1;        }    }