https://pintia.cn/problem-sets/994805342720868352/problems/994805507225665536

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 N​1​​ and N​2​​, 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

代码:

#include <bits/stdc++.h>
using namespace std; string N1, N2;
int radix, tag;
long long sum = 0; long long Pow(long long a, long long b) {
long long ans1 = 1; while(b) {
if(b % 2) {
ans1 = ans1 * a;
b --;
} else {
a = a * a;
b /= 2;
}
}
return ans1;
} long long num(string s, int system) {
int ls = s.length();
reverse(s.begin(), s.end());
long long ans = 0;
if(system <= 10) {
for(int i = 0; i < ls; i ++)
ans += (s[i] - '0') * Pow(system, i);
} else {
int temp;
for(int i = 0; i < ls; i ++) {
if(s[i] >= '0' && s[i] <= '9')
temp = s[i] - '0';
else temp = s[i] - 'a' + 10; ans += temp * Pow(system, i);
}
}
return ans;
} long long Find(string s, long long res) {
char it = *max_element(s.begin(), s.end());
long long l = (isdigit(it) ? it - '0': it - 'a' + 10) + 1;
long long r = max(res, l);
long long mid; while(l <= r) {
mid = (l + r) / 2;
long long rec = num(s, mid);
if(rec == res) return mid;
else if(rec > res || rec < 0) r = mid - 1;
else l = mid + 1;
}
return -1;
} int main() {
cin >> N1 >> N2 >> tag >> radix;
int l1 = N1.length(), l2 = N2.length();
long long out = 0;
if(tag == 1) {
sum = num(N1, radix);
out = Find(N2, sum);
} else {
sum = num(N2, radix);
out = Find(N1, sum);
} if(out == -1) printf("Impossible\n");
else printf("%lld\n", out);
return 0;
}

  明天就过年啦 希望新年会很多不一样 

FHFHFH

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