1010. Radix (25)

//#include<string>
//#include <iomanip>
#include<vector>
#include <algorithm>
//#include<stack>
#include<set>
#include<queue>
#include<map>
//#include<unordered_set>
//#include<unordered_map>
//#include <sstream>
//#include "func.h"
//#include <list>
#include<stdio.h>
#include<iostream>
#include<string>
#include<memory.h>
#include<limits.h>
using namespace std;
int c2int(char c)
{//字符转数字
if (c <= '9'&&c >= '0')
return c - '0';
else if (c <= 'z'&&c >= 'a')
return c - 'a' + 10;
}
int main(void) {

string n1, n2;
int tag;
unsigned long long radix;
cin >> n1 >> n2 >> tag >> radix;
string s, t;
if (tag == 1)
{//tag为1，则radix是n1的进制数
s = n1;
t = n2;
}
else if (tag == 2)
{//tag为2，则radix是n2的进制数
s = n2;
t = n1;
}

unsigned long long sInt = 0;
for (int i = 0; i < s.size(); i++)
{//把s转为十进制
sInt = sInt*radix + c2int(s[i]);
}

unsigned long long minRadix = 2;
for (int i = 0; i < t.size(); i++)
{//求出最小的进制数
minRadix = max(minRadix, (unsigned long long)(c2int(t[i]) + 1));
}
bool flag = false;
unsigned long long result = 0;
unsigned long long r = sInt + 1;//必须+1
unsigned long long l = minRadix;
//使用二分法去查找合适的radix
while (l <= r)//必须要有等号！！
{//从最小的进制数开始遍历
unsigned long long j = (l + r) / 2;//没说明j最大是36进制
unsigned long long tInt = 0;
for (int i = 0; i < t.size(); i++)
{//j为t的进制数，求出t在j进制下的十进制大小
tInt = tInt*j + c2int(t[i]);
if (tInt > sInt)
{//如果尚未统计完，tInt就被sInt大，没必要再统计下去
//说明j进制使得tInt>sInt,t的进制数要比j小
break;
}
}
if (tInt == sInt)
{
flag = true;
result = j;
break;
}
else if (tInt > sInt)
{
r = j - 1;
/*flag = false;
break;*/
}
else l = j + 1;
}
if (flag) cout << result << endl;
else cout << "Impossible" << endl;

return 0;
}