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Input

Output

Examples

Sample Input 1:

Hint

题意:

给出2个数，一个标记t和进制r，对应标记的数为r进制，求两数相等时，另一个数的最小进制

题解:

原本以为是暴力无脑进制转换，后来发现还是有一点小技巧

#include
   
    using namespace std;#define ms(x, n) memset(x,n,sizeof(x));typedef  long long LL;const int INF = 1 << 30;const int MAXN = 1e5+10;LL RTranTo10(int r, string s){
       LL res = 0;    int len = s.length();    for(int i = 0; i < len; ++i){
           if(s[i] >= '0' && s[i] <= '9')            res += (s[i]-'0')*pow(r, len-1-i);        else            res += (s[i]-'a'+10)*pow(r, len-1-i);    }    return res;}int solve(LL n, string s){
       int maxn = 1;    for(int i = 0; i < s.length(); ++i){
           if(isdigit(s[i]))            maxn = max(maxn, s[i]-'0');        else            maxn = max(maxn, s[i]-'a'+10);    }    LL l = maxn+1, r = max(l, n)+1, mid, tmp;    while(l < r){
           mid = (l+r)/2, tmp = RTranTo10(mid, s);        if(tmp < 0 || tmp > n) r = mid; //tmp<0是判断tmp爆longlong溢出了，太坑了        else if(tmp < n) l = mid+1;        else return mid;    }    return -1;}int main() {
       ios::sync_with_stdio(false);    string n1, n2;    int tag, r, ret;    cin >> n1 >> n2 >> tag >> r;    if(tag == 1)        ret = solve(RTranTo10(r, n1), n2);    else        ret = solve(RTranTo10(r, n2), n1);    if(ret == -1)        cout << "Impossible\n";    else        cout << ret << endl;    return 0;}
   

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