PAT 1010

1010. Radix (25)

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 N1 and N2, 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 

刚开始以为此题的范围会超过long long int，试过后发现并没有。另外题目说字符串里面的数字只有{0-9,a-z}，让我以为radix的范围不会超过36，后来发现错了。更改后提交，发现有一个case超时了，估计这个case里的radix非常大，于是radix也采用long long int，并且采用二分搜索，这才解决超时问题。注意这VS下long long int应该定义成__int64，输入输出格式为%I64d。

代码

  1 #define __int64 long long
  2 #include <stdio.h>
  3 #include <string.h>
  4 
  5 long long calculateFunc(char *,long long);
  6 int maxDigit(char *);
  7 int char2num(char);
  8 int main()
  9 {
 10     long long N1,N2,radix;
 11     int tag;
 12     char str1[],str2[];
 13     char *p1,*p2;
 14     while(scanf("%s %s",str1,str2) != EOF){
 15         scanf("%d%I64d",&tag,&radix);
 16         if(tag == ){
 17             p1 = str1;
 18             p2 = str2;
 19         }
 20         else{
 21             p1 = str2;
 22             p2 = str1;
 23         }
 24         N1 = calculateFunc(p1,radix);
 25         int radixMin = maxDigit(p2) + ;
 26         if (radixMin < )
 27             radixMin = ;
 28         int strN = strlen(p2);
 29         if(strN == ){
 30             N2 = char2num(*p2);
 31             if(N2 == N1){
 32                 printf("%I64d\n",N2+);
 33             }
 34             else{
 35                 printf("Impossible\n");
 36             }
 37             continue;
 38         }
 39         long long i = radixMin;
 40         N2 = calculateFunc(p2,i);
 41         int flag = ;
 42         while(N2 <= N1){
 43             if(N1 == N2){
 44                 printf("%I64d\n",i);
 45                 flag = ;
 46                 break;
 47             }
 48             i *= ;
 49             N2 = calculateFunc(p2,i);
 50         }
 51         if(!flag){
 52             long long s = i / ,e = i;
 53             long long mid;
 54               while(e >= s){
 55                 mid = (s + e) / ;
 56                 N2 = calculateFunc(p2,mid);
 57                 if(N1 == N2){
 58                     printf("%I64d\n",mid);
 59                     flag = ;
 60                     break;
 61                 }
 62                 else if (N1 > N2){
 63                     s = mid + ;
 64                 }
 65                 else{
 66                     e = mid - ;
 67                 }
 68               }
 69             if(!flag){
 70                 printf("Impossible\n");
 71             }
 72         }
 73     }
 74     return ;
 75 }
 76 
 77 long long calculateFunc(char *p,long long radix)
 78 {
 79     int n = strlen(p);
 80     int i;
 81     long long N = ;
 82     for(i=;i<n;++i){
 83         N = N * radix + char2num(*(p+i));
 84     }
 85     return N;
 86 }
 87 
 88 int maxDigit(char *p)
 89 {
 90     int n = strlen(p);
 91     int i;
 92     int m = -;
 93     int t;
 94     for(i=;i<n;++i){
 95         t = char2num(*(p+i));
 96         if (t > m)
 97             m = t;
 98     }
 99     return m;
 }
 
 int char2num(char c)
 {
     if(c >= '' && c <= '')
         return c - '';
     else if(c >= 'a' && c <= 'z')
         return c - 'a' + ;
     else
         return -;
 }