PAT A1010 Radix （25 分）——进制转换，二分法

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 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

#include <stdio.h> #include <stdlib.h> #include <string> #include <iostream> #include <math.h> #include <algorithm>
using namespace std; long long str2num(string s, long long radix){ long long num = 0; for (int i = 0; i < s.length(); i++){ if (s[i] >= '0' && s[i] <= '9')num = num + (s[i] - '0')*pow(radix, s.length() - i - 1); else if (s[i] >= 'a' && s[i] <= 'z')num = num + (s[i] - 'a' + 10)*pow(radix, s.length() - i - 1); } return num; } long long min_radix(string s){ long long max_r = 0; for (int i = 0; i < s.length(); i++){ long long tmp; if (s[i] >= '0' && s[i] <= '9')tmp = s[i] - '0'; else if (s[i] >= 'a' && s[i] <= 'z')tmp = s[i] - 'a' + 10; if (tmp>max_r)max_r = tmp; } return max_r+1; } int main(){ string n1, n2; long long num1 = 0, num2 = 0; long long tag, radix; long long min_r = 0, res = 0, le, ri, mid; cin >> n1 >> n2 >> tag >> radix; if (tag == 1){ num1 = str2num(n1, radix); min_r = min_radix(n2); } else{ num1 = str2num(n2, radix); min_r = min_radix(n1); } le = min_r; ri = max(le,num1); while (le <= ri){ mid = (le + ri) / 2; num2 = str2num(tag==1?n2:n1, mid); if (num2<0 || num2 > num1)ri = mid - 1; else if (num2 < num1)le = mid + 1; else{ res = mid; break; } } if (res != 0)printf("%d", res); else printf("Impossible"); system("pause"); }

注意点：第一：两个输入题目保证不超过10位，但没说几进制，因此可能会很大，不过好在测试数据都没有超过long long。

第二：针对这种两个换一换的状况最好写函数，简洁明了，还能够直接用swap函数（algorithm头文件里）

第三：因为没有明确上界，须要本身判断

第四：范围太大，逐个遍历会超时，要用二分法

第五：因为指定数不超过long long，在特定radix下获得的值溢出long long（小于0），说明太大

第六：基数的下限经过遍历字符串获得，下限确定比最大的数大1

第七：上界为ri = max(le,num1);，这里一直想不通，实际上是这样的，若是已知数比最小基数要小，用最小基数也仍是可能算出已知数来的，好比已知10进制的8，未知数为8,9进制就能够了。