1010 Radix 【模拟+二分】

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#模拟 #二分

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本文探讨了如何解决一个有趣的数学问题：给定两个数和其中一个数的进制，如何找出另一个数的进制使得两数相等。通过将已知进制的数转换为十进制，再使用二分查找法寻找未知数的进制，解决了一个可能超过long long数据范围的问题。

题目传送门

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

题目大意：输入两个数a，b，指定其中某个数是几进制数，若要使a=b，那么另一个数应该是几进制数，若有多个答案，输出最小的，若没有，输出Improssible

解题思路：首先将已知进制的数转换为十进制数，然后二分查找另一个数可能的进制数，注意进制可能大于36，数据范围可能超ll

AC代码：

#include <iostream>
#include <map>
using namespace std;
typedef long long ll;
string a,b;
map <char,ll> m;
int judge(ll w,ll val)
{
    ll sum=0;
    for(int i=0;i<b.size();i++){
        sum=sum*w+m[b[i]];
        if(sum>val||sum<0) return 1;//若数据范围超ll，则sum为负值
    }
    if(sum==val) return 0;
    else return -1;
}
int main()
{
    string s,t;
    int flag,radix;
    ll cnt=0;
    for(char i='0';i<='9';i++,cnt++) m[i]=cnt;
    for(char i='a';i<='z';i++,cnt++) m[i]=cnt;
    cin>>s>>t>>flag>>radix;
    if(flag==1) { a=s;b=t; }
    else { a=t;b=s; }
    ll num=0;
    for(int i=0;i<a.size();i++){
        num=num*radix+m[a[i]];
    }
    ll maxx=0;
    for(int i=0;i<b.size();i++){
        maxx=max(maxx,m[b[i]]);
    }
    ll l=maxx+1,r=num+1;
    ll ans=0;
    while(l<=r){
        ll mid=(l+r)>>1;
        int f=judge(mid,num);
        if(f==0) {
            ans=mid;
            break;
        }
        else if(f==1) r=mid-1;
        else l=mid+1;
    }
    if(ans){
        if(b.size()==1) ans=maxx+1;
        cout<<ans<<endl;
    }
    else cout<<"Impossible"<<endl;
    return 0;
}