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

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二分法寻找radix.
二分法中的low与high必须计算得出，low大于N2中出现的最大数。high小于等于N1的值。

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代码1

#include<iostream>
#include<math.h>
#include<string>
using namespace std;
#define maxr 160
string n1,n2,f;
int r;
int getn(char s){
    if(s<='9'&&s>='0')
        return s-'0';
    else return s-'a'+10;
}
long long numr(string s,int r){
    long long ans=0;
    for(int i=0;i<s.size();i++){
        ans=ans*r+getn(s[i]);
    }
    return ans;
}
void find(long long t,long long low,long long high){
    long long  l=low,r=high;
    while(r!=l){
        long long mid=(l+r)/2;
        long long ans=numr(n2,mid);
        if(ans>=t||ans<0){
            r=mid;
        }
        else l=mid+1;
    }
    if(numr(n2,r)!=t){ 
        cout<<"Impossible"<<endl;
    }
    else{
        cout<<r<<endl;
    }
}
int findLargestDigit(string n2){
    int ans=-1,len=n2.size();
    for(int i=0;i<len;i++){
        if(getn(n2[i])>ans){
            ans=getn(n2[i]);
        }
    }
    return ans+1;
}
int main(){
    cin>>n1>>n2>>f>>r;
    if(f=="2"){
        string temp=n1;
        n1=n2;
        n2=temp;
    }
    long long low=findLargestDigit(n2);
    long long t=numr(n1,r);
    long long high=max(low,t)+1;
    find(t,low,high);
    return 0;
}

代码2（晴神）

#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
typedef long long LL;
LL Map[256];
LL inf=(1LL<<63)-1;//long long 最大值为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 converNum10(char a[],LL radix,LL t){
    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;//溢出或者超过N1的十进制
    }
    return ans;
}
int cmp(char N2[],LL radix,LL t){
    int len=strlen(N2);
    LL num=converNum10(N2,radix,t);
    if(num<0) return 1;
    if(t>num) return -1;
    else if(t==num) return 0;
    else return 1;
}
LL binarySearch(char N2[],LL left,LL right,LL t){
    LL mid;
    while(left<=right){
        mid=(left+right)/2;
        int flag=cmp(N2,mid,t);
        if(flag==0) return mid;
        else if(flag==-1) left=mid+1;
        else right=mid-1;
    }
    return -1;
}
int findLargestDigit(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[20],N2[20],temp[20];
int tag,radix;
int main(){
    init();
    scanf("%s %s %d %d",N1,N2,&tag,&radix);
    if(tag==2){
        strcpy(temp,N1);
        strcpy(N1,N2);
        strcpy(N2,temp);
    }
    LL t=converNum10(N1,radix,inf);
    LL low=findLargestDigit(N2);//N2中位数最大的位+1，当成2分下界
    LL high=max(low,t)+1;
    LL ans=binarySearch(N2,low,high,t);
    if(ans==-1) printf("Impossible\n");
    else printf("%lld\n",ans);
    return 0;
}