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：

1. \(a\)、\(b\)、\(c\)是三个正整数；
2. 我们会输入4个乱序的数字：\(x1\)、\(x2\)、\(x3\)、\(x4\)；
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
}

原文地址：https://www.cnblogs.com/xsl19/p/11104948.html