题解【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

简化版题意：

有三个正整数$a$, $b$, $c$, 现在给定$x_{1}$ $=$ $a + b$, $x_{2}$ $=$ $a +c$, $x_{3}$ $=$ $b + c$, $x_{4}$ $=$ $a + b + c$。请求出$a$, $b$, $c$分别是多少。

这是一道数论题。

我们先整理一下题面告诉我们的信息：

$a$、$b$、$c$是三个正整数；
我们会输入4个乱序的数字：$x1$、$x2$、$x3$、$x4$；
$x_{1}$ = $a$ + $b$ , $x_{2}$ = $a$ + $c$ , $x_{3}$ = $b$ + $c$ , $x_{4}$ = $a$ + $b$ + $c$。

∵$a$、$b$、$c$均＞$0$.

∴$x_{4}$是这四个数中最大的数。

至于如何求出$a$、$b$、$c$，则可以：

用$x_{4} - x_{1}$得到$c$，用$x_{4} - x_{2}$得到$b$，用$x_{4} - x_{3}$得到$a$，最后按顺序输出这三个数即可。

注意：我们需要开一个数组$x$[]来存储$x_{1}$、$x_{2}$、$x_{3}$、$x_{4}$，因为这样便于我们排序（可以直接调用$c++$库函数$sort$，但要开头文件$algorithm$），而且可以更好地帮助我们寻找到$a$、$b$、$c$。

Code

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <algorithm>

#include <cmath>//头文件准备

using namespace std;//使用标准名字空间

inline int gi()//快速读入，不解释

{

    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数组的意义同分析

inline void init()//分别输入这四个数

{

	x[1] = gi(), x[2] = gi(), x[3] = gi(), x[4] = gi();

}

inline void solve()//将x数组从小到大排序

{

	sort(x + 1, x + 1 + 4);//1和4是指从x[1]到x[4]从小到大排序

}

inline void output()//输出的自函数

{

	printf("%d %d %d\n", x[4] - x[1], x[4] - x[2], x[4] - x[3]);//分别输出a、b、c。

}

int main()//进入干净整洁的主函数

{

	init();//输入

	solve();//排序解决问题

	output();//输出

	return 0;//养成return 0的好习惯

}