Radix
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
解题思路：
本题给出两个数字n1、n2，给出其中一个数字tag的进制radix，要求判断是否存在某一进制可以使另一个数字与给定数字相等。
为了方便运算，用字符串tn1、tn2记录输入的两个数字，字符串n1记录给定进制的数字，字符串n2记录未确定数字。用map<char, int > mp，记录每个字符所对应的数值，之后，可以先将给定进制的数字n1转化为10进制，n2的进制最小为其包含的最大数字+1记为leftn，且由于n2是整数，所以其进制最大不会超过n1的十进制与leftn中较大的一个+1，记为rightn。以leftn和rightn分别为左右边界二分所有进制，记mid为中点进制，将n2按mid进制转化为10进制与n1的十进制进行比较，如果n2较大证明mid取值过大，将rightn记为mid - 1；若小了，证明mid取值过小，leftn记为mid+1，若正好相等则找到答案。若无法找到某进制使得n1与n2相等，返回-1，如果返回的答案不为-1，输出答案，否则输出Impossible。
AC代码
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 map<char, LL> mp; 5 void init(){ 6 for(char i = '0'; i <= '9'; i++){ 7 mp[i] = i - '0'; //初始化0 - 9 8 } 9 for(char i = 'a'; i <= 'z'; i++){ 10 mp[i] = i - 'a' + 10; //初始化a - z 11 } 12 } 13 LL toDecimal(string a, LL radix, LL maxn){ //转化为10进制的函数，所转化后的数不会超过给出的maxn 14 int len = a.size(); 15 LL ans = 0; 16 for(int i = 0; i < len; i++){ 17 ans = ans * radix + mp[a[i]]; 18 if(ans < 0 || ans > maxn){ //如果数据溢出了或超过上限 19 return -1;//返回-1 20 } 21 } 22 return ans;//返回转化后的值 23 } 24 int cmp(string a, LL radix, LL n1){ //比较函数，用于比较n2的radix进制转化为10进制后与n1的十进制的大小 25 LL n2_10 = toDecimal(a, radix, n1); 26 //获得n2转化为10进制的值 27 if(n2_10 == n1) //如果n2的10进制与n1的十进制相同证明该进制是我们要获得的进制，返回0 28 return 0; 29 else if(n2_10 < 0) //如果toDecimal函数返回的n2小于0，证明n2在该进制下转化为十进制后大于n1的十进制 30 return 1; //进制过大返回1 31 else if(n1 > n2_10) //如果n2在当前进制下转化为10进制小于n1的十进制 32 return -1; //进制过小返回-1 33 else //否则返回1 34 return 1; 35 } 36 LL getRadix(string a, LL leftn, LL rightn, LL n1){ 37 //二分函数传入n2字符串，最小进制，最大进制，n1的十进制值 38 while(leftn <= rightn){ 39 LL mid = (leftn + rightn) / 2; 40 //获得中点 41 LL flag = cmp(a, mid, n1); 42 //判断终点进制n1与n2状态 43 if(flag == 0) //若比较函数返回了0，证明在mid进制下n1与n2相等 44 return mid; //返回mid 45 else if(flag == -1){ //进制过小 46 leftn = mid + 1; 47 }else if(flag == 1){ //进制过大 48 rightn = mid - 1; 49 } 50 } 51 return -1; 52 } 53 int getMaxNum(string a){ //获得n2中最大的数字 54 LL ans = -1; 55 for(string::iterator it = a.begin(); it != a.end(); it++){ 56 ans = max(ans, mp[*it]); 57 } 58 return ans; 59 } 60 int main(){ 61 init(); //初始化mp 62 string tn1, tn2, n1, n2; 63 int tag, radix; 64 cin >> tn1 >> tn2 >> tag >> radix; 65 //输入 tn1 tn2 tag radix; 66 if(tag == 1){ 67 n1 = tn1; 68 n2 = tn2; 69 }else{ 70 n1 = tn2; 71 n2 = tn1; 72 } 73 //n1记录已经确定进制的数字，n2记录未确定的数字 74 LL n1_10 = toDecimal(n1, radix, INT_MAX); 75 //将n1转化为10进制其上限为无穷大 76 LL leftn = getMaxNum(n2) + 1; 77 //获得n2的最小进制 78 LL rightn = max(leftn, n1_10) + 1; 79 //获得n2的最大进制 80 LL ans = getRadix(n2, leftn, rightn, n1_10); 81 //二分所有进制 82 if(tn1 == tn2) 83 printf("%d ", radix); 84 else if(ans == -1){ 85 printf("Impossible "); 86 }else{ 87 printf("%lld ", ans); 88 } 89 return 0; 90 }