1010 Radix (25)

最新推荐文章于 2021-01-19 11:11:55 发布

ChenyutingZJU

最新推荐文章于 2021-01-19 11:11:55 发布

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本文介绍了一道关于进制转换与二分法的编程题。题目要求根据给定的一对正整数和其中一个数的进制，找出另一个数的进制使得两数相等。文章提供了详细的解题思路及C++代码实现。

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1010 Radix (25)
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:

2
Sample Input 2:

1 ab 1 2
Sample Output 2:

Impossible

解题思路：
二分法与进制转换，有点难，看了柳神的解法，能理解，但是让我自己实现就很困难。
convert函数：给定一个数值和一个进制，将它转化为10进制。转化过程中可能产生溢出
find_radix函数：找到令两个数值相等的进制数。在查找的过程中，需要使用二分查找算法，如果使用当前进制转化得到数值比另一个大或者小于0，说明这个进制太大。
还有不知道为什么，我的编译器auto it总是报错，在PAT上是能过，╮(╯▽╰)╭。

#include<cstdio>
#include<string>
#include<iostream>
#include<cctype>
#include<cmath>
#include<algorithm>
using namespace std;
long long convert(string n, long long radix) {
    long long sum = 0;
    int index = 0, temp = 0;
    for (auto it = n.rbegin(); it != n.rend(); it++) {
        temp = isdigit(*it) ? *it - '0' : *it - 'a' + 10;
        sum += temp * pow(radix, index++);
    }
    return sum;
}

long long find(string n,long long num){
    char it = *max_element(n.begin(), n.end());
    long long low = (isdigit(it) ? it - '0': it - 'a' + 10) + 1;
    long long high = max(num, low);
    while (low <= high) {
        long long mid = (low + high) / 2;
        long long t = convert(n, mid);
        if (t < 0 || t > num) high = mid - 1;
        else if (t == num) return mid;
        else low = mid + 1;
    }
    return -1;
}
int main(){
    string n1,n2;
    long long tag,radix,ans;
    cin>>n1>>n2>>tag>>radix;
    if(tag==1){
        ans=find(n2,convert(n1,radix));
    }else{
        ans=find(n1,convert(n2,radix));
    }
    if(ans!=-1) printf("%lld",ans);
    else printf("Impossible");
    return 0;
}