題解【CodeForces1154A】Restoring Three Numbers

Description

Polycarp has guessed three positive integers \(a\), \(b\) and \(c\). He keeps these numbers in secret, but he writes down four numbers on a board in arbitrary order — their pairwise sums (three numbers) and sum of all three numbers (one number). So, there are four numbers on a board in random order: \(a+b\), \(a+c\), \(b+c\) and \(a+b+c\).

You have to guess three numbers \(a\), \(b\) and \(c\) using given numbers. Print three guessed integers in any order.

Pay attention that some given numbers \(a\), \(b\) and \(c\) can be equal (it is also possible that \(a=b=c\)).

Input

The only line of the input contains four positive integers \(x_{1}\),\(x_{2}\),\(x_{3}\),\(x_{4}\)(\(2\)≤\(x_{i}\)≤\(10^{9}\)) — numbers written on a board in random order. It is guaranteed that the answer exists for the given number \(x_{1}\),\(x_{2}\),\(x_{3}\),\(x_{4}\).

Output

Print such positive integers \(a\), \(b\) and \(c\) that four numbers written on a board are values \(a+b\), \(a+c\), \(b+c\) and \(a+b+c\) written in some order. Print \(a\), \(b\) and \(c\) in any order. If there are several answers, you can print any. It is guaranteed that the answer exists.

Examples

Input1:

3 6 5 4

Output1:

2 1 3

Input2:

40 40 40 60

Output2:

20 20 20

Input3:

201 101 101 200

Output3:

1 100 100

Solution

簡化版題意：

有三個正整數\(a\), \(b\), \(c\), 如今給定\(x_{1}\) \(=\) \(a + b\), \(x_{2}\) \(=\) \(a +c\), \(x_{3}\) \(=\) \(b + c\), \(x_{4}\) \(=\) \(a + b + c\)。 請求出\(a\), \(b\), \(c\)分別是多少。

這是一道數論題。

咱們先整理一下題面告訴咱們的信息：

1. \(a\)、\(b\)、\(c\)是三個正整數；

2. 咱們會輸入4個亂序的數字：\(x1\)、\(x2\)、\(x3\)、\(x4\)；

3. \(x_{1}\) = \(a\) + \(b\) , \(x_{2}\) = \(a\) + \(c\) , \(x_{3}\) = \(b\) + \(c\) , \(x_{4}\) = \(a\) + \(b\) + \(c\)。

∵\(a\)、\(b\)、\(c\)均＞\(0\).

∴\(x_{4}\)是這四個數中最大的數。

至於如何求出\(a\)、\(b\)、\(c\)，則能夠：

用\(x_{4} - x_{1}\)獲得\(c\)，用\(x_{4} - x_{2}\)獲得\(b\)，用\(x_{4} - x_{3}\)獲得\(a\)，最後按順序輸出這三個數便可。

注意：咱們須要開一個數組\(x\)[]來存儲\(x_{1}\)、\(x_{2}\)、\(x_{3}\)、\(x_{4}\)，由於這樣便於咱們排序（能夠直接調用\(c++\)庫函數\(sort\)，但要開頭文件\(algorithm\)），並且能夠更好地幫助咱們尋找到\(a\)、\(b\)、\(c\)。

Code

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>//頭文件準備

using namespace std;//使用標準名字空間

inline int gi()//快速讀入，不解釋
{
    int f = 1, x = 0;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
    {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return f * x;
}

int a, b, c, x[5];//a、b、c和x數組的意義同分析

inline void init()//分別輸入這四個數
{
    x[1] = gi(), x[2] = gi(), x[3] = gi(), x[4] = gi();
}

inline void solve()//將x數組從小到大排序
{
    sort(x + 1, x + 1 + 4);//1和4是指從x[1]到x[4]從小到大排序
}

inline void output()//輸出的自函數
{
    printf("%d %d %d\n", x[4] - x[1], x[4] - x[2], x[4] - x[3]);//分別輸出a、b、c。
}

int main()//進入乾淨整潔的主函數
{
    init();//輸入
    solve();//排序解決問題
    output();//輸出
    return 0;//養成return 0的好習慣
}