PAT1010. Radix (25)

题目如下：

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

题意就是给出两个数以及其中一个数的进制，求出另一个数的进制使两个数相等，输出该进制，如果没有该进制则输出Impossible。

看到这个题目，我的想法是首先要写好任意进制转换的函数，以及比较同样进制下两个数的大小。读取数时因为存在英文，则用string来读取。进制比较时则用二分法。也考虑个多个进制都相等的情况，代码大致如下:

#include <iostream>#include <vector>#include <cmath>#include <string>using namespace std;string radixConvert(string number, int oldRadix, int newRadix) //进制转换{string result;int decimal = 0;vector<int> v;//先将该数转换为十进制for (int i = number.size() - 1;i >= 0;i--){if (number[i] >= 'a' && number[i] <= 'z'){v.push_back(number[i] - ('a' - 10));}else{v.push_back(number[i] - ('0' - 0));}}for (int i = 0;i < v.size();i++){decimal += v[i] * pow(oldRadix, i);}//十进制转换为新的进制v.clear();int digit;do {digit = decimal % newRadix;v.push_back(digit);decimal /= newRadix;} while (decimal != 0);for (int i = v.size() - 1; i >= 0;i--){if (v[i] >= 10){result.push_back(v[i] - (10 - 'a'));}else{result.push_back(v[i] - (0 - '0'));}}return result;}int compara(string s1, string s2)//比较字符串里的数字大小{//首先比较位数，位数多的比较大,位数相等时再逐个比较if (s1.size() > s2.size()){return 1;}else if (s1.size() < s2.size()){return -1;}else{if (s1 > s2)return 1;else if (s1 < s2)return -1;elsereturn 0;}}void dichotomy(int low,int high,int radix,string N1,string N2)   //二分法{int mid = (low + high) / 2;while (low <= high){if (compara(N1, radixConvert(N2, mid, radix)) == 1){low = mid + 1;mid = (low + high) / 2;}else if (compara(N1, radixConvert(N2, mid, radix)) == -1){high = mid - 1;mid = (low + high) / 2;}else{cout << mid;break;}}if (low > high)cout << "Impossible";}int findMinRadix(string s)//找出能表示s的最小进制，比如 数字a，那最小进制至少为十一进制{int minRadix;if (s[0] >= 'a' && s[0] <= 'z')minRadix = s[0] - ('a' - 10) + 1;elseminRadix = s[0] - ('0' - 0) + 1;for (int i = 1;i < s.size();i++){if (s[i] >= 'a' && s[i] <= 'z')if (minRadix < s[i] - ('a' - 10) + 1)minRadix = s[i] - ('a' - 10) + 1;elseif(minRadix < s[i] - ('0' - 0) + 1)minRadix = s[i] - ('0' - 0) + 1;}return minRadix;}int main(){string N1, N2,temp;int tag, radix;cin >> N1 >> N2 >> tag >> radix;if (tag == 1){//先处理存在多个进制都使N1和N2相等的情况，这个时候N2只有在最小进制即和N1相等时才会成立if (compara(N1, radixConvert(N2, findMinRadix(N2), radix)) == 0){cout << findMinRadix(N2);}else if (compara(N1,N2) == 1)  //N1和N2同一进制时N1比较大，要相等的话N2的进制比N1的进制大，进制最大为36进制,最小为radix+1，用二分法最快确定进制{dichotomy(radix+1, 36, radix, N1, N2);}else if (compara(N1, N2) == -1)//N1和N2同一进制时N2比较大，要相等的话N2进制比N1的小，最小为1，最大为radix-1{dichotomy(findMinRadix(N2), radix-1, radix, N1, N2);}elsecout << tag;}else{//先处理存在多个进制都使N1和N2相等的情况，这个时候N1只有在最小进制即和N2相等时才会成立if (compara(N2, radixConvert(N1, findMinRadix(N1), radix)) == 0){cout << findMinRadix(N1);}else if (compara(N1, N2) == 1)  //N1和N2同一进制N1大，要相等的话N1的进制比N2的进制小，最大为radix-1,最小为1，用二分法最快确定进制{dichotomy(findMinRadix(N1), radix-1, radix, N2, N1);}else if (compara(N1, N2) == -1)//N1和N2同一进制N1小，要相等的话N1的进制比N2大，最小为radix+1，最大为36{dichotomy(radix+1, 36, radix, N2, N1);}elsecout << tag;}return 0;}

测试用例的结果如下:

去网上查阅了下资料，发现自己对radix的上限设定理解错误了。并不是0-9,a-z总共0-35，radix上限即为36，radix可以很大，只要每位不出现比35大的数字即可表示。

所以重新修改了代码，将radix设为long long类型，对二分法的上下界重新进行了定义，代码如下:

#include <iostream>#include <vector>#include <cmath>#include <string>using namespace std;long long convertToDecimal(string number, long long oldRadix){long long decimal = 0;vector<int> v;//先将该数转换为十进制for (int i = number.size() - 1;i >= 0;i--){if (number[i] >= 'a' && number[i] <= 'z'){v.push_back(number[i] - ('a' - 10));}else{v.push_back(number[i] - ('0' - 0));}}for (int i = 0;i < (int)v.size();i++){decimal += v[i] * pow(oldRadix, i);}return decimal;}string radixConvert(string number, long long oldRadix, long long newRadix) //进制转换{string result;long long decimal = 0;vector<int> v;//先将该数转换为十进制decimal = convertToDecimal(number, oldRadix);//十进制转换为新的进制int digit;do {digit = decimal % newRadix;v.push_back(digit);decimal /= newRadix;} while (decimal != 0);for (int i = v.size() - 1; i >= 0;i--){if (v[i] >= 10){result.push_back(v[i] - (10 - 'a'));}else{result.push_back(v[i] - (0 - '0'));}}return result;}int compara(string s1, string s2)//比较字符串里的数字大小{//首先比较位数，位数多的比较大,位数相等时再逐个比较if (s1.size() > s2.size()){return 1;}else if (s1.size() < s2.size()){return -1;}else{if (s1 > s2)return 1;else if (s1 < s2)return -1;elsereturn 0;}}void dichotomy(long long low,long long high,long long radix,string N1,string N2)   //二分法{long long mid = (low + high) / 2;while (low <= high){if (compara(N1, radixConvert(N2, mid, radix)) == 1){low = mid + 1;mid = (low + high) / 2;}else if (compara(N1, radixConvert(N2, mid, radix)) == -1){high = mid - 1;mid = (low + high) / 2;}else{cout << mid;break;}}if (low > high)cout << "Impossible";}int findMinRadix(string s)//找出能表示s的最小进制，比如 数字a，那最小进制至少为十一进制{int minRadix;if (s[0] >= 'a' && s[0] <= 'z')minRadix = s[0] - ('a' - 10) + 1;elseminRadix = s[0] - ('0' - 0) + 1;for (int i = 1;i < (int)s.size();i++){if (s[i] >= 'a' && s[i] <= 'z')if (minRadix < s[i] - ('a' - 10) + 1)minRadix = s[i] - ('a' - 10) + 1;elseif(minRadix < s[i] - ('0' - 0) + 1)minRadix = s[i] - ('0' - 0) + 1;}return minRadix;}int main(){string N1, N2,temp;int tag;long long radix,decimal;cin >> N1 >> N2 >> tag >> radix;if (tag == 1){//先处理存在多个进制都使N1和N2相等的情况，这个时候N2只有在最小进制即和N1相等时才会成立if (compara(N1, radixConvert(N2, findMinRadix(N2), radix)) == 0){cout << findMinRadix(N2);}else if (compara(N1,N2) == 1)  //N1和N2同一进制时N1比较大，要相等的话N2的进制比N1的进制大，此时二分法下界为N1的进制+1，上界为N1用十进制表示的值+1，用二分法最快确定进制{decimal = convertToDecimal(N1, radix);dichotomy(radix+1, decimal+1, radix, N1, N2);}else if (compara(N1, N2) == -1)//N1和N2同一进制时N2比较大，要相等的话N2进制比N1的小，最小为N2的最小进制，最大为radix-1{dichotomy(findMinRadix(N2), radix-1, radix, N1, N2);}elsecout << tag;}else{//先处理存在多个进制都使N1和N2相等的情况，这个时候N1只有在最小进制即和N2相等时才会成立if (compara(N2, radixConvert(N1, findMinRadix(N1), radix)) == 0){cout << findMinRadix(N1);}else if (compara(N1, N2) == 1)  //N1和N2同一进制N1大，要相等的话N1的进制比N2的进制小，最大为radix-1,最小为N1的最小进制，用二分法最快确定进制{dichotomy(findMinRadix(N1), radix-1, radix, N2, N1);}else if (compara(N1, N2) == -1)//N1和N2同一进制N1小，要相等的话N1的进制比N2大，最小为radix+1，最大为N2的十进制表示+1{decimal = convertToDecimal(N2, radix);dichotomy(radix+1, decimal+1, radix, N2, N1);}elsecout << tag;}return 0;}

此时测试用例情况如下:

又发现自己最后那个应该是输出radix而不是tag，修改代码后用例10通过了，还有3个用例不对。

整体思路应该没问题，可能有些地方还没思考到吧。