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 N?1?? and N?2??, your task is to find the radix of one number while that of the other is given.
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.
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.
6 110 1 10
2
1 ab 1 2
Impossible我本来打算用枚举，这种方法有1个case失误：
#include <iostream> #include <algorithm> using namespace std; /** 一个N 通过radix转换为10进制 */ int itod(string N, int radix) { int p = 1, res = 0; for(int i = N.length() - 1; i >= 0; i--){ if(isdigit(N[i])) res += ((N[i] - ‘0‘) * p); else res += ((N[i] - ‘a‘ + 10) * p); p *= radix; } return res; } int main(){ string N1, N2; int tag, radix; cin >> N1 >> N2 >> tag >> radix; if(tag == 2) swap(N1, N2); // 交换后 radix一定表示N1 int N1d = itod(N1, radix); char max_char = 0;// 求出最大的字符 for(int i = 0; i < N2.length(); i++) if(N2[i] > max_char) max_char = N2[i]; int r;// 最小的进制 if(isdigit(max_char)) r = max_char - ‘0‘ + 1; else r = max_char - ‘a‘ + 11; /** 从最小进制向35进行遍历，如果转换结果相等，则输出 */ for(; r <= 50; r++){ if(itod(N2, r) == N1d) { cout << r; system("pause"); return 0; } } cout << "Impossible"; system("pause"); return 0; }
柳神做法，二分法：
#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; }
原文：https://www.cnblogs.com/littlepage/p/12240223.html