A1010 Radix （25 分）（二分）（难）

最新推荐文章于 2022-03-15 21:45:44 发布

原创最新推荐文章于 2022-03-15 21:45:44 发布 · 185 阅读

0 ·

CC 4.0 BY-SA版权

PAT (Advanced Level) 同时被 2 个专栏收录

84 篇文章

订阅专栏

二分

3 篇文章

订阅专栏

本文探讨了如何解决两个正整数在不同进制下相等的问题，通过算法设计实现进制之间的转换与验证，使用二分搜索法高效求解未知进制。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 tagis 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

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

typedef long long LL;
LL inf = (1LL << 63) - 1;   //long long的最大值2^63-1
LL m[256];      //将0-9，a-z与0-35对应

void init() {
    for(char c = '0'; c <= '9'; c++) {
        m[c] = c - '0';
    }
    for(char c = 'a'; c <= 'z'; c++) {
        m[c] = c - 'a' + 10;
    }
}

LL convertTo10(char a[], LL radix, LL t) {  //将进制为radix的数组转化为10进制，t为上界
    LL ans = 0;
    int len = strlen(a);
    for(int i = 0; i < len; i++) {
        ans = ans * radix + m[a[i]];        //进制转换
        if(ans < 0 || ans > t) {
            return -1;
        }
    }
    return ans;
}

int cmp(char n2[], LL radix, LL t) {        //将N2的十进制与t比较
    int len = strlen(n2);
    LL num = convertTo10(n2, radix, t);
    if(num < 0) {       //N2溢出，肯定是N2 > t
        return 1;
    }
    if(t > num) {       //t比较大，返回-1
        return -1;
    } else if(t == num) {
        return 0;
    } else {
        return 1;
    }
}

LL binarySearch(char n2[], LL left, LL right, LL t) {   //二分求解N2的进制
    LL mid;
    while(left <= right) {
        mid = (left + right) / 2;
        int flag = cmp(n2, mid, t);
        if(flag == 0) {
            return mid;
        } else if(flag == -1) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    return -1;      //解不存在
}

int findLargest(char n2[]) {    //寻找N2的最大数位
    int ans = -1;
    int len = strlen(n2);
    for(int i = 0; i < len; i++) {
        if(m[n2[i]] > ans) {
            ans = m[n2[i]];
        }
    }
    return ans + 1;     //最大数位为ans，说明进制底线为ans+1
}

int main() {
    char n1[20], n2[20], temp[20];
    int tag, radix;
    init();
    scanf("%s %s %d %d", n1, n2, &tag, &radix);
    if(tag == 2) {
        swap(n1, n2);
    }
    LL t = convertTo10(n1, radix, inf);
    LL low = findLargest(n2);   //找到N2的最大数位，当成二分下界
    LL high = max(low, t) + 1;  //上界
    LL ans = binarySearch(n2, low, high, t);    //二分
    if(ans == -1) {
        printf("Impossible\n");
    } else {
        printf("%lld\n", ans);
    }
    return 0;
}