Codeforces Round #552 (Div. 3) —— A. Restoring Three Numbers

A. Restoring Three Numbers

A.恢复三个数字

time limit per test1 second
每次测试的时间限制1秒
memory limit per test256 megabytes
每个测试的内存限制256兆字节
input standard input
输入 标准输入
output standard output
输出 标准输出
Polycarp has guessed three positive integers a, b and c.
Polycarp猜测了三个正整数a、b和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).
他对这些数字保密，但他把四个数字按任意顺序写在黑板上——它们的对和（三个数字）和所有三个数字的和（一个数字）。
So, there are four numbers on a board in random order: a+b, a+c, b+c and a+b+c.
所以，一块电路板上有四个随机排列的数字：A+B，A+C，B+C和A+B+C。

You have to guess three numbers a, b and c using given numbers.
你必须用给定的数字猜测三个数字A、B和C。
Print three guessed integers in any order.
按任意顺序打印三个猜测的整数。

Pay attention that some given numbers a, b and c can be equal (it is also possible that a=b=c).
注意一些给定的数字a、b和c可以相等（也可能是a=b=c）。

Input

输入

The only line of the input contains four positive integers x1,x2,x3,x4 (2≤x_i≤10⁹) — numbers written on a board in random order.
输入的唯一行包含四个正整数x1、x2、x3、x4（2<=x_i<=10^ 9 ^）-以随机顺序写入板上的数字。
It is guaranteed that the answer exists for the given number x1,x2,x3,x4.
保证给定数字x1、x2、x3、x4的答案存在。

Output

输出

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.
打印这样的正整数A、B和C，在一块板上写的四个数字是按某种顺序写的值A+B、A+C、B+C和A+B+C。
Print a, b and c in any order.
按任意顺序打印A、B和C。
If there are several answers, you can print any.
如果有几个答案，你可以打印任何答案。
It is guaranteed that the answer exists.
保证答案存在。

Examples

input
3 6 5 4
output
2 1 3
input
40 40 40 60
output
20 20 20
input
201 101 101 200
output
1 100 100

code

#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
    int num[4];
    cin>>num[0]>>num[1]>>num[2]>>num[3];
    sort(num,num+4);
    cout<<num[3]-num[0]<<' '<<num[3]-num[1]<<' '<<num[3]-num[2]<<endl;
    return 0;
}