程序设计基础51 二分关于进制转换与溢出的处理

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

一，注意点

1，第一个误区是本题中给出了数值的取值范围是0到35我就以为进制的范围就是1到36了，不是的！进制的下界是N2中最小的那一位加1，上界是N1的十进制数值。

2，第二个误区是以为N1的十进制数值不会溢出，实际上是会的，处理方法是若是溢出了返回-1，此时进制的上界变成了0，在二分中很显然就是Impossible.

二，我的代码

#include
#include
#include
using namespace std;
typedef long long LL;
LL INF = (1LL << 63) - 1;
int switch_str_to_num(char a) {
	if (a >= '0'&&a <= '9') {
		return a - '0';
	}
	else {
		return a - 'a' + 10;
	}
}
LL switch_to_radix(char num[], LL radix, LL t) {
	int len = strlen(num);
	LL ans = 0;
	for (int i = 0; i < len; i++) {
		ans = radix*ans + switch_str_to_num(num[i]);
		if (ans < 0||ans > t) {
			return -1;
		}
	}
	return ans;
}
int find_min_radix(char num[]) {
	int min_radix = 0;
	int len = strlen(num);
	for (int i = 0; i < len; i++) {
		if (min_radix < switch_str_to_num(num[i])) {
			min_radix = switch_str_to_num(num[i]);
		}
	}
	return min_radix + 1;
}
LL binary_search(char num[], LL left, LL right, LL x) {
	LL mid;
	if (switch_to_radix(num, left, INF) == x) {
		return left;
	}
	else {
		while (left <= right) {
			mid = (left + right) / 2;
			if (switch_to_radix(num, mid, INF) == x) {
				return mid;
			}
			else if (switch_to_radix(num, mid, INF) > x || switch_to_radix(num, mid, INF) == -1) {
				right = mid - 1;
			}
			else {
				left = mid + 1;
			}
		}
	}
	return -1;
}
int main() {
	char N1[14], N2[14], temp[14];
	int tag = 0, radix_1 = 0, radix_2 = 0;
	LL sum_1 = 0;
	LL lowest_radix = 0, highest_radix = 0;
//	int lowest_radix_1 = 0, lowest_radix_2 = 0, lowest_radix = 0;
	LL ans = 0;
	scanf("%s", N1);
	scanf("%s", N2);
	scanf("%d", &tag);
	scanf("%d", &radix_1);
	if (tag == 2) {
		strcpy(temp, N1);
		strcpy(N1, N2);
		strcpy(N2, temp);
	}
//	lowest_radix_1 = find_min_radix(N1);
	lowest_radix = find_min_radix(N2);
//	lowest_radix = max(lowest_radix_1, lowest_radix_2);
	sum_1 = switch_to_radix(N1, radix_1, INF);
	highest_radix = sum_1 + 1;
//	printf("%lld", ans_1);
	ans = binary_search(N2, lowest_radix, highest_radix, sum_1);
	if (ans != -1) {
		printf("%d", ans);
	}
	else {
		printf("Impossible");
	}
	return 0;
}