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​1​​ and 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

题目大意

题目给出两个数，其中一个的进制已知，求另一个数字的进制使两数大小相等；如该进制不存在，则输出“Impossible”。

解题思路

1. 以字符串形式读取两个数字；
2. 根据Tag及Radix求得其中一个十进制数num（注意：num和需求得的进制都要用long long类型，因为它们的取值可能会很大）；
3. 取目标进制值下限为最大字符代表的十进制数+1，令其为min；目标进制上限为min+1或num+1（取大值）；
4. 通过二分法求得结果，并在过程中剪枝以节约时间；
5. 输出结果并返回0值。

代码

#include<stdio.h>

#define MAXN 15
int ChartoInt(char ch){
    if(ch>='0'&&ch<='9'){
        return ch-'0';
    }else if(ch>='a'&&ch<='z'){
        return ch-'a'+10;
    }else{
        return -1;
    }
}

long long ChartoNum(char ch[],long long radix){
    int i=0;
    long long res=0;
    while(ch[i]!='\0'){
        res=radix*res+ChartoInt(ch[i]);
        i++;
    }
    return res;
}

int FindMin(char a[]){
    int i=0,min=0;
    while(a[i]!=0){
        if(ChartoInt(a[i])>min){
            min=ChartoInt(a[i]);
        }
        i++;
    }
    return min+1;
}

int Compare(long long num,char a[],long long radix){
    long long sum=0;
    int i=0;
    while(a[i]!='\0'){
        sum=sum*radix+ChartoInt(a[i]);
        if(sum>num){
            return 1;
        }
        i++;
    }
    if(sum>num){
        return 1;
    }else if(sum<num){
        return -1;
    }else{
        return 0;
    }
}

int main(){
    char a1[MAXN],a2[MAXN];
    int tag,temp;
    long long radix,num;
    int i=0;
    long long min,max,mid;
    scanf("%s %s ",a1,a2);
    scanf("%d%lld",&tag,&radix);
    if(tag==1){
        num=ChartoNum(a1,radix);
        min=(long long)FindMin(a2);
        max=(num>min?num+1:min+1);
        while(min<max){
            mid=(min+max)/2;
            temp=Compare(num,a2,mid);
            if(temp==1){
                max=mid;
            }else if(temp==-1){
                min=mid;
            }else{
                printf("%d\n",mid);
                return 0;
            }
        }
    }else{
        num=ChartoNum(a2,radix);
        min=(long long)FindMin(a1);
        max=(num>min?num+1:min+1);
        while(min<max){
            mid=(min+max)/2;
            temp=Compare(num,a1,mid);
            if(temp==1){
                max=mid;
            }else if(temp==-1){
                min=mid;
            }else{
                printf("%d\n",mid);
                return 0;
            }
        }
    }
    printf("Impossible\n");
    return 0;
}

运行结果