PAT (Advanced Level) Practice 1010 Radix (25分) (进制转换 longlong溢出 二分查找)

1.题目

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

2.题目分析

开始用正常思路先将一个转为10进制，然后不停转换为其它进制与n2比较，结果测试点 7、10错误

之后得知存在long long溢出以及二分问题，于是改进

https://blog.csdn.net/sheepyWYY/article/details/54576790?depth_1-utm_source=distribute.pc_relevant.none-task-blog-BlogCommendFromBaidu-1&utm_source=distribute.pc_relevant.none-task-blog-BlogCommendFromBaidu-1

3.代码

原版本（测试点7 10 错误）

#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
long long  changea(string c,int m)
{
	long long num = 0;
	for (int i = 0; i < c.length(); i++)
	{
		if (islower(c[i]))  num += (c[i] - 'a' + 10) * pow(m, c.length() - 1 - i); //如果是字母的话就+10，因为a代表10
		else if (isupper(c[i])) num += (c[i] - 'A' + 10)*pow(m, c.length() - 1 - i);
		else if (isdigit(c[i]))num += (c[i] - '0')*pow(m, c.length() - 1 - i);//如果是数字就直接乘
	}
	return num;
}
string changeb(long long  num,int n)
{
	string answer = "";
	int out[100];
	int s = 0;
	while (num > 0)
	{
		out[s++] = num%n;
		num /= n;
	}
	if (n <= 9)//将变为的进制<=9
	{
		for (int i = s - 1; i >= 0; i--)
			answer += (out[i] + '0');
	}
	else
	{
		for (int i = s - 1; i >= 0; i--)
		{
			if (out[i] >= 10)answer += (out[i] - 10 + 'a'); 
			else answer += (out[i] + '0');
		}
	}
	return answer;
}

int main()
{
	string n1, n2;
	long long tag, radix;
	cin >> n1 >> n2 >> tag >> radix;
	if (tag == 2)
		swap(n1, n2);
	long long num = changea(n1, radix);

	for (int i = 2; i < 10000; i++)
	{
		string temp = changeb(num, i);
		if (temp == n2) { printf("%lld", i); return 0; }
	}
	printf("Impossible\n");
}

改进：

#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

int Alpha2int(char c) { //数字转换
	if (int(c) >= int('0') && int(c) <= int('9')) {
		return (int(c) - 48);
	}
	else
		return int(c) - 87;
}
long long  changea(string c,int m)
{
	long long num = 0;
	for (int i = 0; i < c.length(); i++)
	{
		if (islower(c[i]))  num += (c[i] - 'a' + 10) * pow(m, c.length() - 1 - i); //如果是字母的话就+10，因为a代表10
		else if (isupper(c[i])) num += (c[i] - 'A' + 10)*pow(m, c.length() - 1 - i);
		else if (isdigit(c[i]))num += (c[i] - '0')*pow(m, c.length() - 1 - i);//如果是数字就直接乘
		if (num < 0)return -1;
	}
	return num;
}
long long changeb(long long  start,long long end,long long num ,string n2)
{
	long long mid = (start + end) / 2;
	long long temp = changea(n2, mid);
	long long result = -1; 
	if (start <= end) 
	{
		if (temp == -1 || temp > num) 
		{//溢出减少Radix
			result = changeb(start, mid - 1, num, n2);
		}
		else if (temp < num)
		{
			result = changeb(mid + 1, end, num, n2);
		}
		else 
		{
			result = mid;
		}
	}
	return result;
}

int main()
{
	string n1, n2;
	long long tag, radix;
	cin >> n1 >> n2 >> tag >> radix;
	if (tag == 2)
		swap(n1, n2);
	long long num = changea(n1, radix);
	long long max = num + 1;
	long long min = 2;
	for (int i = 0; i < n2.size(); i++) 
	{ 
		if (Alpha2int(n2[i]) >= min)
			min = Alpha2int(n2[i]) + 1; //要加1
	}
		long long result = changeb(min, max,num,n2);
		if (result !=-1) { printf("%lld", result); return 0; }
	printf("Impossible\n");
}