PTA-1010——Radix（未完全通过，通过率21/25）

题目:

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

题目分析:

二分查找，进制转换。中间有很多坑，具体见代码中的注释。参考博客（但还有少量数据出错，没找到错在哪）

代码:

 1 #include<iostream>
 2 using namespace std;
 3 string n1,n2;
 4 int tag,radix;
 5 int main(){
 6     cin>>n1>>n2>>tag>>radix;
 7     long long x=0;    //此题进制转换后可能超过int的范围，要用long long类型 
 8     if(tag==2){
 9         string temp=n1;
10         n1=n2;
11         n2=temp;
12     }
13     
14     //将第一个数字转为10进制
15     int t=1;;
16     for(int i=n1.length()-1;i>=0;i--){
17         if(n1[i]>='0'&&n1[i]<='9'){
18             x+=t*(n1[i]-'0');
19         }else{
20             x+=t*(n1[i]-'a'+10);
21         }
22         t*=radix;
23     }
24     
25     //二分查找，否则容易超时，且题目没有说明最大的进制是36 
26     long long l,r;    //二分查找法的上下界 
27     l=1; 
28     for(int i=0;i<n2.length();i++){        //下界的最小值 
29         if(n2[i]>='0'&&n2[i]<='9'){
30             if(n2[i]-'0'>l){
31                 l=n2[i]-'0';
32             }
33         }else{
34             if(n2[i]-'a'+10>l){
35                 l=n2[i]-'a'+10;
36             }
37         } 
38     } 
39     l++;
40     if(x>l){    //如6 110,此时x=6,l=2,满足x>l，即r=7，因为如果比7更大，那么这个数的十进制最小也要比7大 
41         r=x+1;
42     }else{        //如3 45，此时x=3,l=6,满足x<l,即r=7 
43         r=l+1;
44     }
45     
46     //将第二个数字转为10进制 
47     bool flag=false;
48     long long y;
49     while(l<=r){
50         long long mid=(l+r)/2;
51         y=0;
52         t=1;
53         for(int i=n2.length()-1;i>=0;i--){
54             if(n2[i]>='0'&&n2[i]<='9'){
55                 y+=t*(n2[i]-'0');
56             }else{
57                 y+=t*(n2[i]-'a'+10);
58             }
59             t*=mid;
60         }
61         if(y==x){
62             cout<<mid;
63             flag=true;
64             break;
65         }else if(y<0||y>x){        //考虑y<0是因为：可能出现转换后溢出的情况，这时候y会小于0 
66             r=mid-1;
67         }else if(y<x){
68             l=mid+1;
69         }
70     }
71     if(!flag){
72         cout<<"Impossible";
73     }
74     return 0;
75 }