[CodeForces-1225A] Forgetting Things 【构造】

[CodeForces-1225A] Forgetting Things 【构造】

Tags: problem solution codeforces problem solution construction

---

Title Description

Time limit
2000 ms
Memory limit
524288 kB
Source
Technocup 2020 - Elimination Round 2
Tags
math *900
Site
https://codeforces.com/problemset/problem/1225/a

Face questions

Example
Input1

1 2

Output1

199 200

Input2

4 4

Output2

412 413

Input3

5 7

Output3

-1

Input4

6 2

Output4

-1

Subject to the effect

Given (d_a, d_b \) \ number as a (first from left) on the highest of the two numbers, I asked if he could find such a \ (A, b \) , so that \ (a + 1 = b \ ) . If the output is not -1.

For example,
given 12, we can easily find 199 + 1 = 200, then we output 199,200.
Given 62, we could not find \ (a, b \) to meet the requirements, then we output -1.

---

Resolve

Simple construction.

Method of constructing the answer should have a variety.

Points discuss,
if given two numbers are equal, then output \ (D_A \ Times 10 \; \; D_B \ Times + 10. 1 \) .
If a given \ (. 1 + D_A D_B = \) , then the direct output \ (d_a \; \; d_b \) can.
A special case, if a given \ (D_A = 9. 1 D_B = \) , then the output \ (9, 10 \) .
Other cases, no solution, then output \ (--1 \) .

---

By Code

/*
Status
    Accepted
Time
    31ms
Memory
    12kB
Length
    309
Lang
    GNU G++11 5.1.0
RemoteRunId
    67064706
*/

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

int main()
{
    int x, y;

    cin >> x >> y;

    if(x == y)
        cout << x * 10 << " " << y * 10 + 1;
    else if(x + 1 == y)
        cout << x << " " << y;
    else if(x == 9 && y == 1)
        cout << "9 10";
    else
        cout << -1;

    return 0;
}

---