PATA 1010 Radix（进制转换）

1010 Radix

分数 25

作者 CHEN, Yue

单位 浙江大学

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

注意另外一个字符串的进制并不是简单的最大36进制，有可能是无穷进制，这就需要用二分法来求解。并且需要把两外一个字符串的最小进制求解出来，否则会出错（我写了另外一个代码，没有求出最小进制，有一个测试点过不了，现在也没找出原因）。并且转进制所求的数应该开 long long ,int 会爆。

AC代码：

 
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
 
using namespace std;
 
typedef long long LL;
 
//val初值为无穷大，得到第一个十进制数，看stoLL函数就能理解了
LL val = 1e18; 
 
//s radix进制转换为十进制
LL stoLL(string s, LL radix)
{
    LL res = 0, index = 1;
    for(int i=s.size()-1; i>=0; --i)
    {
        int v;
        if(isdigit(s[i]))
            v = s[i] - '0';
        else
            v = s[i] - 'a' + 10;
        
        res += v * index;
        index *= radix;
        if(res > val) //如果res大于val，说明radix进制肯定大了
            break;
    }
     
    return res; 
}
 
int main()
{
    string s[3];
    LL tag, radix;
    
    cin >> s[1] >> s[2] >> tag >> radix;
    
    val = stoLL(s[tag], radix);
    
    //找到另外一个字符串的最小进制
    LL l = 0, r = val + 1;
    for(int i=0;i<s[tag ^ 3].size();++i)
    {
        LL v;
        if(isdigit(s[tag ^ 3][i]))
            v = s[tag ^ 3][i] - '0';
        else
            v = s[tag ^ 3][i] - 'a' + 10;
        l = max(l,v + 1);
    }
    
    //二分
    while(l < r)
    {
        LL mid = l + r >> 1;
        if(stoLL(s[tag ^ 3], mid) >= val)
            r = mid;
        else
            l = mid + 1;
    }
    
    if(stoLL(s[tag ^ 3], l) == val)
        cout << l << endl;
    else
        puts("Impossible");
    
    return 0;
}

有一个测试点通不过，求大佬找错误：

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
 
using namespace std;
 
typedef long long LL;
 
LL val = 1e18;
 
LL stoLL(string s, LL radix)
{
    LL res = 0, index = 1;
    for(int i=s.size()-1; i>=0; --i)
    {
        int v;
        if(isdigit(s[i]))
            v = s[i] - '0';
        else
            v = s[i] - 'a' + 10;
            
        //如果radix比本身的元素都要小，那么这个进制肯定小了，将答案赋为负值退出
        if(v >= radix) 
        {
            res = -1e9;
            break;
        }
        res += v * index;
        index *= radix;
        if(res > val) 
            break;
    }
      
    return res; 
}
 
int main()
{
    string s[3];
    LL tag, radix;
    
    cin >> s[1] >> s[2] >> tag >> radix;
    
    val = stoLL(s[tag], radix);
    
    LL l = 0,r = max(val + 1, 36ll);
    
    while(l < r)
    {
        LL mid = l + r >> 1;
        if(stoLL(s[tag ^ 3], mid) >= val)
            r = mid;
        else
            l = mid + 1;
    }
    
    if(stoLL(s[tag ^ 3], l) == val)
        cout << l << endl;
    else
        puts("Impossible");
    
    return 0;
}