PAT甲級1010 （進位制和二分法）
題目
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
題解： 一開始不瞭解tag的意思，看到if那句話，明白了它只能是1或2。radix就是N tag的進位制數。一開始想的是，把未知進位制的數轉化成和已知進位制數相同的進位制，但這樣操作難度很高，不如全轉化為10進位制。
本題還有一個重點是，並不會因為每一位取0~35而使得輸入數最大為36進位制，進位制可能超大，所以用long long radix。
未知進位制數，最小進位制是它最大的數字+1，因為要和已知進位制的數相等，所以最大進位制是已知進位制數的進位制。假設例子裡N2為6進位制的010，化為10進位制就是6=N1。進位制數再大化為10進位制一定超過6了。
#include <iostream>
#include <cctype>
#include <algorithm>
#include <cmath>
using namespace std;
long long convert(string n, long long radix) {
long long sum = 0;
int index = 0, temp = 0;
for (auto it = n.rbegin(); it != n.rend(); it++) {
temp = isdigit(*it) ? *it - '0' : *it - 'a' + 10;
sum += temp * pow(radix, index++);
}
return sum;
}
long long find_radix(string n, long long num) {
char it = *max_element(n.begin(), n.end());
long long low = (isdigit(it) ? it - '0': it - 'a' + 10) + 1;
long long high = max(num, low);
while (low <= high) {
long long mid = (low + high) / 2;
long long t = convert(n, mid);
if (t < 0 || t > num) high = mid - 1;
else if (t == num) return mid;
else low = mid + 1;
}
return -1;
}
int main() {
string n1, n2;
long long tag = 0, radix = 0, result_radix;
cin >> n1 >> n2 >> tag >> radix;
result_radix = tag == 1 ? find_radix(n2, convert(n1, radix)) : find_radix(n1, convert(n2, radix));
if (result_radix != -1) {
printf("%lld", result_radix);
} else {
printf("Impossible");
}
return 0;
}