1010. Radix (25)

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本文介绍了一个进制转换问题，即给定两个数和其中一个数的进制，如何找到另一个数的正确进制使得两个数相等。文章通过二分搜索算法高效地解决了这一问题，并详细解释了算法的实现细节。

1010. Radix (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 N1 and N2, 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:

Sample Input 2:

1 ab 1 2

Sample Output 2:

Impossible

提交代码

题意: 给定两个数, 其中一个数已经知道了它的进制表示, 求另一个数用何进制表示才能同其相同, 如果不能, 输出Impossible.

首先给定一组数据 "zzzzzzzzzz ab 1 36" 如果普通模拟肯定超时, 所以我们采用二分答案的思想来做. 好了, 大体方向已经找到了, 那么二分的上下限呢, 下限我们可以很快求出, 就是这个数中最大的那个数+1, 可是上限呢? 我们可以求出这个数的长度n, 假设, 我们要以R进制表示, 且另外一个数是X, 则, 原来的数 < R + R^1 + R^2 + ...... R^n = (R^n - 1) / (R - 1), 则 (R^n - 1) / (R - 1) = X, R ^ n = (X * (R - 1) + 1), lg(X * (R - 1) + 1) / lg (R) = n, 我们粗略可以得出 (lgX / lgR) + 1 = n.至此, 上限就可以求出来啦.

#include <cstdio>
#include <cstring>
#include <cmath>

typedef long long LL;
const LL INF = 0x3f3f3f3f3f3f3f3f;

LL get_digit(char ch) {
	if (ch >= '0' && ch <= '9')
		return ch - '0';
	return ch - 'a' + 10;
}

LL find_x(char* str, LL radix) {
	int len = strlen(str);
	LL ans = get_digit(str[len - 1]);
	len = len - 2;
	LL tmp = radix;

	while (~len) {
		ans += get_digit(str[len]) * tmp;
		tmp *= radix;
		--len;
	}

	return ans;
}

LL search(char* str, LL a, LL R) {
	LL left = R, right;
	if (strlen(str) > 1)
		right = exp(log(a) / (strlen(str) - 1));
	else
		right = a + 1;

	while (left <= right) {
		LL mid = (left + right) / 2;
		LL tmp = find_x(str, mid);
		if (tmp == a)
			return mid;
		else if (tmp < a)
			left = mid + 1;
		else
			right = mid - 1;
	}

	return -1;
}

int main() {
	LL a, b, tag, radix;

	char stra[20], strb[20];
	scanf("%s%s%lld%lld", stra, strb, &tag, &radix);

	if (tag == 2) {
		char tmp[20];
		strcpy(tmp, stra);
		strcpy(stra, strb);
		strcpy(strb, tmp);
	}

	a = find_x(stra, radix);

	LL R = 2;
	for (int i = 0; strb[i]; ++i)
		if (R <= get_digit(strb[i]))
			R = get_digit(strb[i]) + 1;

	R = search(strb, a, R);

	if (R == -1)
		printf("Impossible");
	else
		printf("%lld", R);

	return 0;
}