1010 Radix（25 分）

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 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

//本题关键是确定进制的上限，可以有很大很大，故而设为long long类型
//进制转换和二分查找法是本题的解决方法

#include<stdio.h>
#include<iostream>
#include<string>

using namespace std;

//字符转化为数字
int charToInt(char ch)
{
    int rs= 0;
    if(ch >= ‘0‘ && ch <= ‘9‘)
    {
        rs = ch - ‘0‘;
    }

    if(ch >= ‘a‘ && ch <= ‘z‘)
    {
        rs = ch - ‘a‘ + 10;
    }

    return rs;
}

//化为十进制数
long long changeToDecimal(string s,long long radix)
{
    long long result = 0;
    for(int i = 0;i < s.length();i++)
    {
        result = result*radix + charToInt(s[i]);
    }

    return result;
}

//比较n1与n2在十进制下的大小，此函数可避免运行超时
int compare(long long n1, string n2, long long radix)
{
    long long sum = 0;
    for (int i = 0; i < n2.length();i++)
    {
        sum = sum * radix + charToInt(n2[i]);
        if (sum > n1) //重点，避免运行超时
        {
            return 1;
        }
    }
    if (sum > n1)// n1 < n2
    {
        return 1;
    }
    else if (sum < n1) // n1 > n2
    {
        return -1;
    }
    else //相等
    {
        return 0;
    }

}

//二分查找，十进制数与未知进制数相等时的进制
long long Binary_Search(long long n,string p)
{
    //step1:未知进制数的进制下界（每个位置数字的最大值+1）
    //      进制上界（数d1和数b最大符号代表的数的最大值加上1,n>最小进制时，则为n+1；否则为最小进制，可以设为low+1）

    int max = 0;
    for(int i = 0;i < p.length();i++)//从末位开始
    {
        if(max < charToInt(p[i])) //不为空
        {
            max = charToInt(p[i]);
        }
    }

    long long low = (long long)max + 1;//记录最小的进制；
    long long high;//记录最大进制；
    if(n > low)
    {
        high = n + 1;
    }
    else
    {
        high = low + 1;
    }
    long long mid;//中间数

    //step2:二分查找
    while(low < high)
    {
        mid = (low + high)/2;
        if(compare(n,p,mid) == 0)//(n == changeToDecimal(p,mid))
        {
            return mid;
        }
        else if(compare(n,p,mid) == 1)//(n < changeToDecimal(p,mid))
        {
            high = mid;   //while中是low <= high时，为mid-1；此处是low < high；但按照书中一模一样的写法，只能得24分
        }
        else
        {
            low = mid;    //while中是low <= high时，为mid+1；此处是low < high
        }
    }

    return -1;
}

int main()
{
    string N1,N2;
    int tag;
    long long radix;
    cin >> N1;
    cin >> N2;
    cin >> tag;
    cin >> radix;

    //全部化为10进制比较
    long long d1,d2;//N1和N2化为十进制的值
    if(tag == 1)
    {
        d1 = changeToDecimal(N1,radix);
        if(Binary_Search(d1,N2) == -1)
        {
            cout << "Impossible" << endl;
        }
        else
        {
            cout << Binary_Search(d1,N2) << endl;
        }
    }

    if(tag == 2)
    {
        d2 = changeToDecimal(N2,radix);
        if(Binary_Search(d2,N1) == -1)
        {
            cout << "Impossible" << endl;
        }
        else
        {
            cout << Binary_Search(d2,N1) << endl;
        }
    }

    return 0;
}

原文地址：https://www.cnblogs.com/meiqin970126/p/9607252.html