PAT A1010

题目：https://pintia.cn/problem-sets/994805342720868352/problems/994805507225665536

注意点：数据范围大，要用LL，并且防止加减溢出。用高效的搜索方法。不注意这些只能得23分。因为有一个测试用例是LL范围的，还有一个如果从2开始递增验证，要超时。

#include <cstdio>
#include <string>
#include <iostream>
using namespace std;
typedef long long LL;
const LL INF = (1LL<<63)-1; //INF
int tonum(char a){
	if(a >= '0' && a <= '9')
		return a - '0';
	else
		return a -'a' + 10;
}
LL todecimal(string str, LL radix){
	LL num = 0;
	for(int i = 0; i < str.length(); i++){
		int a = tonum(str[i]);
		if(a >= radix)
			return -1;
		num = num * radix + a;
		if(num < 0)        //溢出
			return INF;
	}
	return num;
}
LL binarysearch(LL l, LL x, string str){
	LL r;
	if(x > 36)
		r = x;           //最大进制为另一个数
	else
		r = 36;
	while(l < r){
		LL mid = l/2 + r/2;
		LL a = todecimal(str, mid);
		if(a >= x){
			r = mid;
		}else
			l = mid + 1;
	}
	return l;
}
int main(){
	string n1, n2;
	int tag;
	LL radix;
	cin>>n1>>n2>>tag>>radix;
	LL a;
	if(tag == 1){
		a = todecimal(n1, radix);
		LL ans = binarysearch( 2, a, n2);
		if(a == todecimal(n2, ans))
			printf("%lld", ans);
		else
			printf("Impossible");
	}else{
		a = todecimal(n2, radix);
		LL ans = binarysearch( 2, a, n1);
		if(a == todecimal(n1, ans))
			printf("%lld", ans);
		else
			printf("Impossible");
	}
	return 0;
}