PAT甲级1010 Radix (25)

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 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

这题陷阱很多- -，首先想到的是把已给出进制的那个数转化为10进制，然后另外一个未知进制的数去一个个试可能的进制，然后也转换为10进制和那个已知的比较，看看是否存在一个进制使他们相等，提醒，进制会非常非常大，要用long long存。首先是进制的范围（举个例子）：

比如6 110 1 10

这里6已经是10进制了，我们看110，最小的进制是2，那么最大的进制不能超过6+1=7，因为如果进制为8，那么110转换成十进制为9必定比6大，已经无意义了。

#include
#include
#include
using namespace std;

char n1[15], n2[15];
int t, r;

long long toDecimal(char s[],long long r) {
	long long res = 0;
	long long temp = 0;
	int len = strlen(s);
	for (int i = 0; i < len; i++) {
		if (s[i]>='0'&&s[i]<='9') {
			temp = s[i] - '0';
		}
		else if (s[i] >= 'a'&&s[i] <= 'z') {
			temp = s[i] - 'a' + 10;
		}
		res = res * r + temp;
	}
	return res;
}

long long binary(long long l, long long r,long long target) {
	while (l <= r) {
		long long mid = (l + r) / 2;
		if (toDecimal(n2, mid) == target) {
			return mid;
		}
		else if (toDecimal(n2, mid) < 0 || toDecimal(n2, mid) > target) {
			r = mid - 1;
		}
		else if (toDecimal(n2, mid) < target) {
			l = mid + 1;
		}
	}
	return -1;
}

long long getMin() {
	int min = 0,tmp = 0;
	for (int i = 0; i < strlen(n2);i++) {
		if (n2[i] >= 'a'&&n2[i] <= 'z') {
			tmp = n2[i] - 'a' + 10;
		}
		else if('0' <= n2[i] && n2[i] <= '9'){
			tmp = n2[i] - '0';
		}
		if (tmp > min) min = tmp + 1;
	}
	if (min < 2) return 2;
	return min;
}

int main() {
	long long res = 0, radix = 0;
	scanf("%s %s %d %d", n1, n2, &t, &r);

	if (t == 2) swap(n1, n2);
	res = toDecimal(n1, r);

	long long left = getMin();//比如2进制只有0和1，那么最低进制就是该字符串最大数字+1
	long long right = max(left, res) + 1;//最大进制不能超过res+1，因为超过了肯定比res大
	radix = binary(left, right, res);

	if (radix != -1) printf("%lld\n", radix);
	else printf("Impossible");

	return 0;
}