PAT 1010 Radix

原创已于 2022-03-12 13:41:03 修改 · 298 阅读

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#c++ #pat考试

于 2022-03-12 13:34:19 首次发布

这篇博客讨论了一种解决正整数对N1和N2在不同基数下相等的问题。给出的C++代码实现了一个算法，该算法找到使等式N1 = N2成立的另一个数的基数，当给定基数时。关键点包括处理溢出、二分搜索的边界以及在多个解中选择最小基数。

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.

踩坑点：

① 一开始没注意到有多个满足的进制时，选最小的进制。

②注意二分法的上界不是36，应该是基准数转换为十进制加1！比如10 10000 0 10，输出应该是10000.（必须要加1，这点还没弄明白）

③注意二分搜索时left right以及最后输出的radix类型都要设置成long long

④转化为10进制时，由于radix是long long，可能会发生溢出，溢出判断：<0，此时radix应该减小

#include<cstdio>
#include<iostream>
#include<math.h>
#include<string.h>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<ctime>

using namespace std;

struct Bign {
	int d[15];
	int len;
	long long  radix;
	Bign() {
		memset(d, 0, sizeof(d));
		len = 0;
	}
};
Bign num_to_arr(string str) {
	Bign temp;
	int index = 0;
	temp.len = str.length();
	for (int i = 0; i < temp.len; i++) {
		if (str[i] >= '0' && str[i] <= '9')
			temp.d[temp.len - 1 - i] = str[i] - '0';
		else
			temp.d[temp.len - 1 - i] = str[i] - 'a'+10;
	}
	return temp;
}
long long arr_to_num(Bign a) {
	int len = a.len;
	long long r = a.radix;
	long long m=0;
	for (int i = len-1; i >=0; i--) {
		m = m * r + a.d[i];
		if (m < 0)
			return -1;
	}
	return m;
}

long long find_radix(Bign a, long long m,long long left, long long right) {
	
	while (left <right) {
		long long mid = (left + right) / 2;
		a.radix = mid;
		long long k = arr_to_num(a);
        if(k>=m||k==-1){
			right = mid;
		}
		else
			left = mid + 1;
	}
	return left;
}
int main() {
	string N1, N2;
	int flag;
	long long radix;
	cin >> N1;
	cin >> N2;
	scanf("%d %lld", &flag, &radix);
	Bign a = num_to_arr(N1);
	Bign b = num_to_arr(N2);
	long long m;
	long long  k;
	if (flag == 1) {
		a.radix = radix;
		m = arr_to_num(a);
		int max = b.d[0];
		for (int i = 0; i < b.len; i++) {
			if (b.d[i] > max)
				max = b.d[i];
		}
		k = find_radix(b, m, max + 1, m+1);
		b.radix = k;
		long long t = arr_to_num(b);
		if(t!=m)
			printf("Impossible");

		else
			printf("%d", k);

	}
	else {
		b.radix = radix;
		m = arr_to_num(b);
		int max = a.d[0];
		for (int i = 0; i < a.len; i++) {
			if (a.d[i] > max)
				max = a.d[i];
		}
		k = find_radix(a, m, max + 1, m+1);

		a.radix = k;
		long long t = arr_to_num(a);
		if (t != m)
			printf("Impossible");

		else
			printf("%d", k);
	}


	return 0;
}