1010. Radix (25)

本文介绍了一个进制转换问题,即给定两个正整数和其中一个数的进制,需要找出另一个数的正确进制,使得两个数相等。文章提供了详细的代码实现,包括如何将不同进制的数转换为十进制数,以及如何通过二分查找确定未知进制。

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1010. Radix (25)
时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B
判题程序 Standard 作者 CHEN, Yue

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

注意点:

  • 进制范围要进行限制,为最大位+1 至 题目给出进制对应的数据的十进制值加1
  • 可能溢出,所以要特殊处理
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstring>
#include <fstream>
#include <string>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
using namespace std;

const int MaxN = 100010;

int Map[130];

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



long long StringToLongLong(string a,long long radix)
{
    if (radix == 0) return 0;
    long long res = 0;
    long long POW = 1;
    int size = a.size();
    for (int i = size - 1; i >= 0; --i, POW *= radix)
        res += Map[a[i]] * POW;
    if (res < 0)
        return -1;
    return res;
}

int FindLow(string a)
{
    int size = a.size();
    int max = -1;
    for (int i = 0; i < size; ++i)
    {
        if (Map[a[i]] > max)
            max = Map[a[i]];
    }
    return max + 1;
}


bool CanbeDone(string n,long long & radix,long long value,long long left,long long right)
{
    long long val, mid;
    while (left <= right)
    {
        mid = left + (right - left) / 2;
        val = StringToLongLong(n, mid);
        if (value < val || val == -1)
            right = mid - 1;
        else if (value > val)
            left = mid + 1;
        else
        {
            radix = mid;
            return true;
        }

    }
    return false;
}


int main()
{
#ifdef _DEBUG
    freopen("data.txt", "r+", stdin);
    //fstream cin("data.txt");
#endif // _DEBUG

    string N1, N2;
    int tag;
    long long num1,maxradix, radix;

    init();

    cin >> N1 >> N2 >> tag >> radix;
    if (tag == 2)
        swap(N1, N2);

    num1 = StringToLongLong(N1, radix);
    maxradix = num1 + 1;

    long long low = FindLow(N2);
    if (CanbeDone(N2, radix, num1, low, max(low,maxradix)))
        printf("%lld", radix);
    else
        printf("Impossible");


#ifdef _DEBUG
    //cin.close();
#ifndef _CODEBLOCKS
    std::system("pause");
#endif // !_CODEBLOCKS
#endif // _DEBUG

    return 0;
}
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