pat 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

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:

Sample Input 2:

1 ab 1 2

Sample Output 2:

Impossible


思路：将给定的数转换成十进制数,然后对另一个数字用循环逐个找到对应进制的十进制数的大小
　　　本来用stoi函数的，写的很方便，但是通过牛客给的测试数据发现，stoi的进制范围是有要求的，2-36进制的准换
难点：（1）进制范围的确定，下界 l 等于待测的数字的各个位上的最大的数+1,上届是给定数字的十进制形式+1
         (2) 要用long long 以及对于代码中溢出的处理   ，，，，long long 溢出后小于零
疑问：题中要求如果答案不唯一，那么输出最小的，但是感觉二分来写会满足不了题意，但是最后ac了

代码如下：

 1 #include <cstdio>
 2 #include <string>
 3 #include <iostream>
 4 #include <algorithm>
 5 #define LL long long
 6 using namespace std;
 7 
 8  LL ch_num(string s,int r){
 9     LL res=0;
10     int len=s.length();
11     for(int i=0;i<len;i++){
12         if(s[i]>='0'&&s[i]<='9')
13             res=res*r+(s[i]-'0');
14         else
15             res=res*r+(s[i]-'W');
16     }
17     if(res<0)
18         return -1;//溢出处理
19     return res;
20 }
21 LL maxcnt(string s){
22     LL maxn=0,t;
23     int len=s.length();
24     for(int i=0;i<len;i++){
25         if(s[i]>='0'&&s[i]<='9'){
26             t=s[i]-'0';
27             maxn=max(maxn,t);
28         }
29         else{
30             t=s[i]-'W';
31             maxn=max(maxn,t);
32         }
33     }
34     return maxn+1;
35 }
36 
37 int main(void)
38 {
39     int tag,rad;
40     string fin,sen;
41     cin>>fin>>sen>>tag>>rad;
42 
43     LL num,l;
44     string str;
45     if(tag==1){
46         num=ch_num(fin,rad);
47         l=maxcnt(sen);
48         str=sen;
49     }
50     else{
51         num=ch_num(sen,rad);
52         l=maxcnt(fin);
53         str=fin;
54     }
55 
56     LL mid,tem;
57     LL r=num+1;
58     while(l<=r){
59         mid=(l+r)>>1;
60         tem=ch_num(str,mid);
61         if(tem==num){
62             printf("%lld\n",mid);
63             return 0;
64         }
65         if(tem>num || tem==-1)
66             r=mid-1;
67         else
68             l=mid+1;
69     }
70     printf("Impossible\n");
71 
72     return 0;
73 }