題目大意:
- 給定2個數字num1和num2,和num1的進制,要求求出在num1==num2的條件下num2的進制,如果無法滿足,返回impossible.
注意:
- 思想
- 將num1轉化爲10進制。
- num2的進制有一個區間,low是num2的每個數位+1,high是num1+1
- 用二分法尋找num2的進制。 - 進制涉及到溢出問題,所有數據類型設置爲long long (超出範圍視作impossible)
- num可能是字母,a~z分別映射爲10—35,用到anto迭代器和isdigit(it)函數
- 尋找num2每個數位的最大數字,可以用到*max_element(n.begin(), n.end());
#include <iostream>
#include <cctype>
#include <algorithm>
#include <cmath>
using namespace std;
long long convert(string n, long long radix)
{
long long sum = 0;
int index = 0, temp = 0;
for (auto it = n.rbegin(); it != n.rend(); it++)
{
temp = isdigit(*it) ? *it - '0' : *it - 'a' + 10;
sum += temp * pow(radix, index++);
}
return sum;
}
long long find_radix(string n, long long num)
{
char it = *max_element(n.begin(), n.end());
long long low = (isdigit(it) ? it - '0' : it - 'a' + 10) + 1;
long long high = max(num, low);
while (low <= high)
{
long long mid = (low + high) / 2;
long long t = convert(n, mid);
if (t < 0 || t > num)
high = mid - 1;
else if (t == num)
return mid;
else
low = mid + 1;
}
return -1;
}
int main()
{
string n1, n2;
long long tag = 0, radix = 0, result_radix;
cin >> n1 >> n2 >> tag >> radix;
result_radix = tag == 1 ? find_radix(n2, convert(n1, radix)) : find_radix(n1, convert(n2, radix));
if (result_radix != -1)
{
printf("%lld", result_radix);
}
else
{
printf("Impossible");
}
return 0;
}