1010. Radix (25分)
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 结尾无空行
Sample Output 1:
2 结尾无空行
Sample Input 2:
1 ab 1 2 结尾无空行
Sample Output 2:
Impossible 结尾无空行
#include <iostream> #include <cmath> #include <algorithm> using namespace std; typedef long long LL; // 变为10进制数 LL convert(string str, LL radix) { LL len = str.length(), decimal = 0; for (LL i = 0; i < len; i++) { LL n = isdigit(str[i]) ? str[i] - '0' : str[i] - 'a' + 10; decimal += n * pow(radix, len - i - 1); } return decimal; } // 二分查找 LL find_radix(string str, LL n1) { char n = *max_element(str.begin(), str.end()); LL left = isdigit(n) ? n - '0' + 1 : n - 'a' + 11; LL right = n1 + 1; while (left <= right) { LL mid = (left + right) / 2; LL n2 = convert(str, mid); if (n2 < 0 || n2 > n1) { right = mid - 1; } else if (n2 == n1) { return mid; } else { left = mid + 1; } } return -1; } int main() { string s1, s2; LL tag, radix; cin >> s1 >> s2 >> tag >> radix; if (tag == 2) { swap(s1, s2); } LL n1 = convert(s1, radix); LL rst = find_radix(s2, n1); if (rst != -1) { printf("%lld", rst); } else { printf("Impossible"); } return 0; }
先把已知进制数转化为十进制，然后未知进制数用不同进制转十进制代入比较，这里使用的是二分查找的方法进行优化。
