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

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
long long stringToInteger(char* s,int radix){
    long long result=0;
    int i=(int)strlen(s)-1;
    long long product=1;
    while(i>=0){
        long long tmp=0;
        if(s[i]>='0'&&s[i]<='9'){
            tmp=(s[i]-'0');
        } else if(s[i]>='a'&&s[i]<='z'){
            tmp=(s[i]-'a'+10);
        }
        result+=tmp*product;
        if(result<0){//溢出
            return -1;
        }
        product*=radix;
        i--;
    }
    return result;
}
int main(){
    freopen("in.txt","r",stdin);
    char n1[20];
    char n2[20];
    int tag,radix;
    scanf("%s %s %d %d",n1,n2,&tag,&radix);
    if(tag==2){
        swap(n1,n2);
    }
    long long target=stringToInteger(n1,radix);
    long long maxRadix=target+1;//最大的进制
    long long minRadix=0;
    //找出下界
    for(int i=0;n2[i]!='\0';i++){
        if(n2[i]>='0'&&n2[i]<='9'){
            minRadix=minRadix<(n2[i]-'0'+1)?(n2[i]-'0'+1):minRadix;
        }else if(n2[i]>='a'&&n2[i]<='z'){
            minRadix=minRadix<(n2[i]-'a'+11)?(n2[i]-'a'+11):minRadix;
        }
    }
    long long mid,end_radix;
    while(minRadix<=maxRadix){
        mid=minRadix+(maxRadix-minRadix)/2;
        long long result=stringToInteger(n2,mid);
        if(result==-1||result>target){//进制太大
            maxRadix=mid-1;
        } else if(result<target){
            minRadix=mid+1;
        } else {
            printf("%d",mid);
            return 0;
        }
    }
    printf("Impossible");
    return 0;
}

这里是参考了别人的代码。简述下思路，首先将其中一个数转换为10进制数，转换后的数+1是上界，下界是n2中最大的那一项+1。然后因为范围很大，所以要二分查找。在转换10进制数时，要考虑进制过大时，long long可能会溢出。