标签：strong temp sdi span string mil html git hat
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.
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.
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.
6 110 1 10
2
1 ab 1 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
标签：strong temp sdi span string mil html git hat
原文地址：https://www.cnblogs.com/zlrrrr/p/10350969.html
