1010 Radix (25 point(s)) - C语言 PAT 甲级

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分类专栏： PAT 甲级文章标签： PAT 甲级 C

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本文介绍了一个算法挑战，涉及两个正整数N1和N2的基数转换和比较。任务是在给定一个数的基数和另一个数的情况下，找到使等式成立的未知数的基数。文章详细解释了输入和输出规格，提供了示例输入和输出，并分享了一个C语言实现的解决方案，使用二分查找法来高效地确定未知基数。

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1010 Radix (25 point(s))

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:

6 110 1 10

Sample Output:

2

题目大意：

N1 和 N2 各自不超过10位，如果 tag 为 1，则最后一个数字 radix 为 N1 的基数，如果标记为 2，则为 N2 的基数。数字 N1 和 N2 小于其基数，并从集合 {0-9，a-z} 中选择，其中 0-9 表示十进制数字 0-9，a-z 表示十进制数字 10-35。

设计思路：

两个数均转化为 10 进制数进行比较
寻找另一个数的基数时：
- 确定基数的最小值和最大值
- 二分法查找基数
当另一个数只有一位时，在任何大于它本身的基数下，其值均等于它自身：
- 如果等于已知数，最小进制为自身加 1
- 否则无解

编译器：C (gcc)

#include <stdio.h>
#include <ctype.h>
#include <limits.h>
#include <string.h>

#define CBASE10(C) ((C) >= '0' && (C) <= '9' ? (C) - '0' : (C) - 'a' + 10)

long long convert10(char *s, long long radix);
long long minradix(char *s);
long long binsearch(char *s, long long n, long long rmin, long long rmax);

int main(void)
{
        long long tag, radix;
        char n1[11], n2[11], *s1, *s2;
        long long r, m1, rmin, rmax;

        scanf("%s %s %d %d", n1, n2, &tag, &radix);
        if (tag == 1) {
                s1 = n1;
                s2 = n2;
        } else {
                s1 = n2;
                s2 = n1;
        }

        m1 = convert10(s1, radix);
        rmin = minradix(s2);
        rmax = LLONG_MAX;
        if (strlen(s2) == 1) {
                if (m1 == rmin - 1)
                        printf("%lld", rmin);
                else
                        printf("Impossible");
        } else {
                r = binsearch(s2, m1, rmin, rmax);
                if (r != -1)
                        printf("%lld", r);
                else
                        printf("Impossible");
        }

        return 0;
}

long long convert10(char *s, long long radix)
{
        long long n, sum;
        for (sum = 0; *s; s++) {
                n = CBASE10(*s);
                if((LLONG_MAX - n) / radix < sum)
                        return -1;
                sum = sum * radix + n;
        }
        return sum;
}

long long minradix(char *s)
{
        char r;
        long long n;
        for (r = '0'; *s; s++)
                if (*s > r)
                        r = *s;
        return CBASE10(r) + 1;
}

long long binsearch(char *s, long long n, long long rmin, long long rmax)
{
        long long r, m;
        while (rmax >= rmin) {
                r = rmin + (rmax - rmin) / 2;
                if ((m = convert10(s, r)) > n || m == -1)
                        rmax = r - 1;
                else if (m < n)
                        rmin = r + 1;
                else
                        return r;
        }
        return -1;
}