1010 Radix (25)（25 point(s)）

problem

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

tip

給出兩個數以及其中一個數的進制，求兩個數相等時另外一個數可能的進制。（數只可能由0-1，a-z組成，但進制可能會超過36進制，坑點。）

answer

#include<bits/stdc++.h>
using namespace std;

#define INF 0x3f3f3f3f

long long  tag, radix;
string a, b;

void Impossible(){
    cout<<"Impossible";
}

long long  GetIndex(char a){
    return isdigit(a) ? a - '0' : a - 'a' + 10;
}

long long  CalStringToInt(string s, long long  index){
    long long  a = 0;
    int i = 0;
    string::reverse_iterator it;
    for(it = s.rbegin(); it != s.rend(); it++){
        int temp = isdigit(*it) ? *it - '0' : *it - 'a' + 10;
        a += temp * pow((float)index, i++);
    }
    return a;
}

int main(){
//  freopen("test.txt", "r", stdin);
    cin>>a>>b>>tag>>radix;
    long long  maxIndex= 0;
    string A;
    long long  goal;
    if(tag == 1) {
        A = b;
        goal = CalStringToInt(a, radix);
    }else {
        A = a;
        goal = CalStringToInt(b, radix);
    }
    char me = *max_element(A.begin(), A.end());
    maxIndex = (isdigit(me)?me-'0':me-'a'+10) +1;
    
    long long left = maxIndex, right = max(maxIndex, goal), mid = (left+right)/2;
    bool flag = false;
    while(left <= right){
        mid = (left+right)/2;
        long long temp = CalStringToInt(A, mid);
        if(temp < 0 || temp > goal) right = mid -1;
        else if(temp == goal) {flag = true; break;}
        else left = mid +1;
    }
    if(flag) cout<<mid; else Impossible();
    return 0;
}

experience

• c++:
  • max_element 函數

// min_element/max_element example

#include <iostream>     // std::cout
#include <algorithm>    // std::min_element, std::max_element

bool myfn(int i, int j) { return i<j; }

struct myclass {
  bool operator() (int i,int j) { return i<j; }
} myobj;

int main () {
  int myints[] = {3,7,2,5,6,4,9};

  // using default comparison:
  std::cout << "The smallest element is " << *std::min_element(myints,myints+7) << '\n';
  std::cout << "The largest element is "  << *std::max_element(myints,myints+7) << '\n';

  // using function myfn as comp:
  std::cout << "The smallest element is " << *std::min_element(myints,myints+7,myfn) << '\n';
  std::cout << "The largest element is "  << *std::max_element(myints,myints+7,myfn) << '\n';

  // using object myobj as comp:
  std::cout << "The smallest element is " << *std::min_element(myints,myints+7,myobj) << '\n';
  std::cout << "The largest element is "  << *std::max_element(myints,myints+7,myobj) << '\n';

  return 0;
}

- string interator && reverse_interator

// string::begin/end
#include <iostream>
#include <string>

int main ()
{
  std::string str ("Test string");
  for ( std::string::iterator it=str.begin(); it!=str.end(); ++it)
    std::cout << *it;
  std::cout << '\n';
  for (std::string::reverse_iterator rit=str.rbegin(); rit!=str.rend(); ++rit)
    std::cout << *rit;
  return 0;
}