1010 Radix (25 分) 超级坑恶魔坑

1010 Radix (25 分)

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:

Sample Input 2:

1 ab 1 2

Sample Output 2:

Impossible

具体代码

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <math.h>
using namespace std;
typedef long long LL;
LL Map[256];
LL inf=pow(2,63)-1;void init(){for(char c='0';c<='9';c++){Map[c]=c-'0';}for(char c='a';c<='z';c++){Map[c]=c-'a'+10;}
}LL num10(char a[],LL radix,LL t){//将a转换为10进制，t为上界LL ans=0;int len=strlen(a);for(int i=0;i<len;i++){ans=ans*radix+Map[a[i]];if(ans<0||ans>t)return -1;}return ans;}int cmp(char N2[],LL radix,LL t){int len=strlen(N2);LL num=num10(N2,radix,t);if(num<0)return 1;//溢出，N2>t;if(t>num)return -1;//t较大，返回-1else if(t==num)return 0;//相等else return 1;//num较大
}LL binary(char N2[],LL left,LL right,LL t){LL mid;while(left<=right){mid=(left+right)/2;int flag=cmp(N2,mid,t);//N2转为10进制后与tj比较if(flag==0)return mid;else if(flag==-1)left=mid+1;else right=mid-1;}return -1;
}int find(char N2[]){int ans=-1,len=strlen(N2);for(int i=0;i<len;i++){if(Map[N2[i]]>ans){ans=Map[N2[i]];}}return ans+1;
}char N1[15],N2[15],temp[15];
int tag,radix;
int main(){init();scanf("%s %s %d %d",N1,N2,&tag,&radix);if(tag==2){swap(N1,N2);}
LL t=num10(N1,radix,inf);//将N1z转换为10进制
LL low=find(N2);//找到N2中最大的数位
LL high=max(low,t)+1;//上界
LL ans=binary(N2,low,high,t);
if(ans==-1)printf("Impossible\n");
else printf("%lld\n",ans);
return 0;}