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 N?1?? and N?2??, 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 radixHere 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输入： 两个数N1，N2, tag(1:radix是N1的基数, 2:radix 是N2的基数), radix
输出：若两个数可互相转化 则另一个数的基数 ，
若不能互相转换 则输出 Impossible
转：https://blog.csdn.net/liuchuo/article/details/52495925
分析：convert函数：给定一个数值和一个进制，将它转化为10进制。转化过程中可能产生溢出
find_radix函数：找到令两个数值相等的进制数。在查找的过程中，需要使用二分查找算法，如果使用当前进制转化得到数值比另一个大或者小于0，说明这个进制太大～
#include <iostream>
#include <cctype>
#include <algorithm>#include <cmath>
using namespace std;
long long convert(string n, long long radix) {//将已知进制的数转换成十进制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());//二分查找，找到令两个数相等的基数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; //进制太大else if (t == num) return mid;else low = mid + 1; //进制太小 }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;
}
auto关键字是c++11新引入的关键字,自动推断出变量的类型
rbegin() 返回一个逆序迭代器，它指向容器c的最后一个元素
rend() 返回一个逆序迭代器，它指向容器c的第一个元素前面的位置
min_element()和max_element
头文件：#include<algorithm>
作用：返回容器中最小值和最大值。max_element(first,end,cmp);其中cmp为可选择参数!