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

我的方法就是遍历所有进制。做完发现其实用一个二位数组写就不用把相同代码写两遍了

最后错一个点，一直没找出来。百度，看的都是二分法，估计是超时了吧

#include<iostream>#include<cstring>#include<cstdio>#include<queue>#include<stack>#include<algorithm>#include<cmath>using namespace std;int main(){char num1[100],num2[100];int flag;int radix;cin>>num1>>num2>>flag>>radix;if(flag==1){long long sum=0;for(int i=0;i<strlen(num1);i++){if(num1[i]>='0'&&num1[i]<='9'){sum=sum*radix+num1[i]-'0';}else{sum=sum*radix+num1[i]-'a'+10;}}int max=1;for(int i=0;i<strlen(num2);i++){if(num2[i]>='0'&&num2[i]<='9'){if(num2[i]-'0'>max) max=num2[i]-'0';}else{if(num2[i]-'a'+10>max) max=num2[i]-'a'+10;}} int f=0;for(int i=max+1;i<=100000;i++){long long sum2=0;for(int j=0;j<strlen(num2);j++){if(num2[j]>='0'&&num2[j]<='9'){sum2=sum2*i+num2[j]-'0';if(sum2>sum) break;    }    else{sum2=sum2*i+num2[j]-'a'+10;if(sum2>sum) break;    }}if(sum2==sum){f=1;cout<<i<<endl;break;}}if(f==0){cout<<"Impossible";}}else{    long long sum=0;for(int i=0;i<strlen(num2);i++){if(num2[i]>='0'&&num2[i]<='9'){sum=sum*radix+num2[i]-'0';}else{sum=sum*radix+num2[i]-'a'+10;}}int max=1;for(int i=0;i<strlen(num1);i++){if(num1[i]>='0'&&num1[i]<='9'){if(num1[i]-'0'>max) max=num1[i]-'0';}else{if(num1[i]-'a'+10>max) max=num1[i]-'a'+10;}} int f=0;for(int i=max+1;i<=100000;i++){long long sum2=0;for(int j=0;j<strlen(num1);j++){if(num1[j]>='0'&&num1[j]<='9'){sum2=sum2*i+num1[j]-'0';if(sum2>sum) break;    }    else{sum2=sum2*i+num1[j]-'a'+10;if(sum2>sum) break;    }}if(sum2==sum){f=1;cout<<i<<endl;break;}}if(f==0){cout<<"Impossible";}} return 0;}