PAT 1010 Radix

个人学习记录，代码难免不尽人意

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 N 1and N 2, 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

#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const ll inf=(1ll << 63)-1; 
int chartoint(char ch){
	if(ch>='0'&&ch<='9') return ch-'0';
	else return ch-'a'+10;
}
ll converttoten(string str,ll radix){
	ll num=0;
	for(int i=0;i<str.size();i++){
		num=num*radix+chartoint(str[i]);
	}
	if(num<0||num>inf) return -1;
	return num;
}
int cmp(string n2,ll radix,ll num1){
	ll num2=converttoten(n2,radix);

	if(num2<0) return 1;
	if(num1>num2) return -1;
	if(num1==num2) 
		return 0;
	if(num1<num2) return 1;
}
ll binarysearch(string n2,ll left,ll right,ll num1){
	ll mid;
	while(left<=right){
		mid=(left+right)/2;
		int flag=cmp(n2,mid,num1);
		if(flag==1) right=mid-1;
		else if(flag==-1) left=mid+1;
		else return mid;
	}
	return -1;
}
int main(){
   int tag;
   ll radix;
   string n1,n2;
   cin >> n1 >>n2 >> tag >> radix;
   if(tag!=1){
   	string temp=n1;
   	n1=n2;
   	n2=temp;
   }

   ll num1=0,num2=0;
   num1=converttoten(n1,radix);
   ll low=1;
   for(int i=0;i<n2.size();i++)
      {
      	if(chartoint(n2[i])>low) low=chartoint(n2[i]);
	  }
	low++;
    ll upper=max(low,num1)+1;
    ll res=binarysearch(n2,low,upper,num1);
    if(res==-1) printf("Impossible\n");
    else printf("%lld\n",res);
   return 0;
  
}

这道题很有难度，建议参考《算法笔记》。