【PAT - 甲级1010】Radix (25分)（二分，进制转化）

题干：

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

题目大意：

给定两个字符串，每个都是0~9 , a~z 代表0~35这36个数字。

给定一个数字的radix，你的任务是找到另一个数字的基数，使得N1=N2。

解题报告：

我们知道，把十进制数转化成其他进制是困难的，但是把其他进制转化成十进制是较为简单的。所以直接锁定一个目标进制之后，转化成十进制比较两个数字式是否相同就可以了。刚开始想错了，以为就是最多就是35进制了，但是其实则不然，可以是无穷进制，所以不能直接枚举了。考虑到上界是比较好确定的，并且随着radix的增大，转化的十进制数是单调递增的，考虑二分。

注意二分的上下界限制就好了。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
char n1[MAX],n2[MAX],ned[MAX];
int tag,radix;
__int128 shi;
int go(char c) {
	if(c >= '0' &&  c <= '9') return c - '0';
	else return c - 'a' + 10;
}
__int128 trans(char s[],ll rdx) {//把s从rdx进制转化成10进制 
	__int128 res = 0;
	int len = strlen(s);
	for(int i = 0; i<len; i++) {
		res = res * rdx + go(s[i]);
		if(res < 0) return (__int128)9e18 * 2;
	}
	return res;
}

int main()
{
	cin>>n1>>n2>>tag>>radix;
	
	if(tag == 1) shi = trans(n1,radix),strcpy(ned,n2);
	else shi = trans(n2,radix),strcpy(ned,n1);
	
	int mx = 0,len = strlen(ned);
	for(int i = 0; i<len; i++) {
		mx = max(mx,go(ned[i]));
	}
	__int128 l = mx+1,r = 9e18,mid;
	ll ans=-1;
	while(l<=r) {
		mid = (l+r)>>1;
		__int128 tmp = trans(ned,mid); 
		if(tmp < shi) l = mid+1;
		else r = mid-1;
		if(tmp == shi) ans = mid;
	}
	if(ans != -1) printf("%lld\n",ans);
	else printf("Impossible\n");
	return 0 ;
}