1010 Radix

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

题意：给出两个数，求在已知其中一个数的进制下，另外一个数最小进制下两个数相等。

解析：误导信息：the set { 0-9, a-z } where 0-9 represent the decimal numbers 0-9, and a-z represent the decimal numbers 10-35. 以为进制最大也就36 其实是可以无穷大。但是也不是，在测试数据中，超过long long就是溢出，则Impossible。可知一个个遍历过去肯定不理想，超时。可以想到二分查找。之前一直在第七个测试点卡着只有24分，现在终于明白，原来若存在进制，这最大也就是da，因为超过da，不可能da==db。

#include<bits/stdc++.h>
using namespace std;#define e exp(1)
#define pi acos(-1)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define ll long long
#define ull unsigned long long
#define mem(a,b) memset(a,b,sizeof(a))
int gcd(int a,int b){return b?gcd(b,a%b):a;}int main()
{char a[15],b[15];int c[15];ll tag,radix,l,la;ll da=0,db=0;cin>>a>>b>>tag>>radix;if(tag==1){ll k=1;la=strlen(a);for(ll i=la-1; i>=0; i--){if(a[i]>='a'&&a[i]<='z'){da+=(a[i]-'a'+10)*k;}else da+=(a[i]-'0')*k;k*=radix;}l=strlen(b);for(ll i=l-1; i>=0; i--){if(b[i]>='a'&&b[i]<='z'){c[i]=b[i]-'a'+10;}else c[i]=b[i]-'0';}}if(tag==2){ll k=1;la=strlen(b);for(ll i=la-1; i>=0; i--){if(b[i]>='a'&&b[i]<='z'){da+=(b[i]-'a'+10)*k;}else da+=(b[i]-'0')*k;k*=radix;}l=strlen(a);for(int i=l-1; i>=0; i--){if(a[i]>='a'&&a[i]<='z'){c[i]=a[i]-'a'+10;}else c[i]=a[i]-'0';}}ll maxx=0;for(ll i=0; i<l; i++){if(c[i]>maxx)maxx=c[i];}ll left=maxx+1,right=da+1;//最大的进制ll ans;int f=0;while(left<=right){ll mid=(left+right)/2;db=0;ll k=1;for(ll i=l-1; i>=0; i--){db+=c[i]*k;k*=mid;}if(da==db){ans=mid;f=1;right=mid-1;}else if(db<0||db>da){right=mid-1;}else if(db<da){left=mid+1;}}if(!f)cout<<"Impossible"<<endl;elsecout<<left<<endl;return 0;
}

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