PAT甲级1010 Radix （25 分)【二分查找+进制转换】

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

题意：进制转换，给出2个数，后面跟着tag，和radix的值，tag表示第几个数的进制是radix，要找出另外一个数的进制能和这个radix进制的数相等。题目没给出进制的范围，所以需要用到long long类型。

解题思路：这题单纯模拟有4个测试点过不了，所以扣了4分，然后通过看别人博客的解题思路，才知道要用二分法查找，当时没多想就顺序搜索，看了他们的思路，将自己的代码改了一遍。就增加了一个二分查找函数find_radix(string s,int num)，将当前的进制和另外一个转换的数比较，如果当前进制比另外一个大，就说明这个进制太大了，但在转换的时候可能会溢出，所以小于0也说明太大了，这时将high=mid-1，如果等于当前的就直接返回这个mid，如果是其他情况，就说明比较小，将low=mid+1，具体见代码

#include<bits/stdc++.h>
using namespace std;
string n1,n2;
long long radix;
long long change(string s,long long r)
{
	int n=s.length(),k=0,sum=0;
	for(int i=n-1;i>=0;i--)
	{ if(isdigit(s[i])) {
		sum+=(s[i]-'0')*pow(r,k);
    }else if(isalpha(s[i]))
		sum+=(s[i]-'W')*pow(r,k);
		k++;
	}
	return sum;
}
long long find_radix(string n,long long num) 
{
	char m=*(max_element(n.begin(),n.end()));
	long long low=(isdigit(m)?m-'0':m-'a'+10)+1;
	long long high=max(low,num);
	while(low<=high)
	{
		long long mid=(high+low)/2;
		long long sum=change(n,mid);
		if(sum<0||sum>num) high=mid-1;
		else if(sum==num) return mid;
		else  low=mid+1;
	}
	return -1;
}
int main(void)
{
	int tag;
	cin>>n1>>n2;
	cin>>tag>>radix;
	string n,an;
	long long ans;
	if(tag==1) n=n1,an=n2;
	else if(tag==2) n=n2,an=n1;
	ans=find_radix(an,change(n,radix));
	if(ans==-1)printf("Impossible");
	else printf("%lld",ans);
	return 0;
	
}