PAT-1010 Radix

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 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

算法说明：

已知一个数和其基数，求另一个数的基数使得这两个数相等。数字表示使用[0-9a-z]，很容易看出基数的范围是[2-36]，如果已知数很大，基数是会超出36的，可能会很大很大，用long long 来存储这个基数，如果你用暴力遍历查找基数，时间会超时，可以用二分查找。

#include <iostream>
#include <cstring>
using namespace std;
#define Max 3
// 转成十进制
long long toDecimal(char *N,long long radix){
	long long decimal=0;
	for(int i=0;i<strlen(N);i++){
		if(N[i]<58)
			decimal=decimal*radix+(N[i]-48);
		else
			decimal=decimal*radix+(N[i]-87); // a-z ,10-35 呵呵
	}
	return decimal;
}
//找出串中最大值
int maxValueStr(char *N){
	int max=0;
	for(int i=0;i<strlen(N);i++)
		if(N[i]<58&&N[i]-48>max)
			max=N[i]-48;
		else if(N[i]-87>max)
			max=N[i]-87;
	if(max>1)
		return max;
	else
		return 1;
}
int compare(char *N,long long radix,long long target){
	long long decimal=0;
	for(int i=0;i<strlen(N);i++){
		if(N[i]<58)
			decimal=decimal*radix+(N[i]-48);
		else
			decimal=decimal*radix+(N[i]-87);
		if(decimal>target||decimal<0)
			return 1;
	}
	if(decimal>target)
		return 1;
	else if(decimal<target)
		return -1;
	else
		return 0;
}
long long binarySearch(long long target,char *N,long long low,long long high){
	long long mid=low;

	while(low<=high){
		if(compare(N,mid,target)==1)
			high=mid-1;
		else if(compare(N,mid,target)==-1)
			low=mid+1;
		else
			return mid;
		mid=(low+high)/2;
	}
	return -1;
}
int main(int argc, char* argv[])
{
	char *N[Max];

	long long radix,target;
	int tag;
	for(int i=0;i<Max;i++)
		N[i]=new char[10]();

	cin >> N[1] >> N[2] >> tag >> radix;

	target=toDecimal(N[tag],radix); //目标数转成十进制比较
	tag=(tag==1)? 2:1;
	long long max=(long long)maxValueStr(N[tag])+1;  // 出现最大值f 进制为16 +1 

	if(target<=max){
		long long result=binarySearch(target,N[tag],max,36);
		if(result==-1)
			cout<<"Impossible";
		else
			cout<<result;
	}else{
		long long result=binarySearch(target,N[tag],max,target+1);
		if(result==-1)
			cout<<"Impossible";
		else
			cout<<result;
	}
	return 0;
}

2020考研打卡第十三天，星辰之变，骄阳岂是终点。

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