URI 1241 Fit Or Don't Fit | Solution in C++

URI Online Judge | 1241

Fit or Dont Fit II

By Neilor Tonin, URI  Brazil

Timelimit: 1

Paulinho have again in your hands another problem. Now the teacher asked him to make a new program to verify from two big numbers A and B, if B corresponds to the last digit of A.

Input

The input consists of several test cases. The first line of input contains an integer N that indicates the number of test cases. Each test case consists of two numbers A and B greather than zero, with up to 1000 digits each.

Output

For each test case, print a message informing if the second number fit ("encaixa" in portughese) or didn't fit ("nao encaixa") in the first number.

Sample Input	Sample Output
4 56234523485723854755454545478690 78690 5434554 543 1243 1243 54 64545454545454545454545454545454554	encaixa nao encaixa encaixa nao encaixa

URI 1241 Fit Or Don't Fit | Solution in C++

#include <cstdio>

#include <iostream>

#include <vector>

#include <string>

using namespace std;

int main()

{

    int n, sizeA, sizeB, tmp;

    string a, b;

    bool check;

    scanf("%i", &n);

    for (int i = 0; i < n; ++i)

    {

        check = true;

        cin >> a >> b;

        sizeA = a.length(); sizeB = b.length();

        char arrayA[sizeA + 1]; char arrayB[sizeB + 1];

        if(sizeA < sizeB){

            puts("nao encaixa");

            continue;

        }

        for (int i = 0; i < sizeA; ++i)

            arrayA[i] = a.at(i);

             

        for (int i = 0; i < sizeB; ++i)

            arrayB[i] = b.at(i);

        tmp = sizeA - sizeB;

        for (int i = 0; tmp < sizeA; ++tmp, ++i)

        {

            if(arrayA[tmp] != arrayB[i]){

                check = false;

                break;

            }

        }

        if(check == true){

            puts("encaixa");

        }else{

            puts("nao encaixa");

        }

    }

    return 0;

}

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