一、Restoring Three Numbers

本题链接：Restoring Three Numbers

题目：
A. Restoring Three Numbers
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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 x1,x2,x3,x4 (2≤xi≤1e9) — numbers written on a board in random order. It is guaranteed that the answer exists for the given number x1,x2,x3,x4.

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
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

本博客给出本题截图：

题意：给出四个数，这些数分别是a + b，a + c，b + c，a + b + c的结果，求a b c

AC代码

#include <iostream>
#include <algorithm>

using namespace std;

int main()
{
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    
    if (a > b) swap(a, b);
    if (b > c) swap(b, c);
    if (c > d) swap(c, d);
    
    int x = (a - c + b) / 2;
    int y = (a + c - b) / 2;
    int z = (b + c - a) / 2;
    
    cout << x << ' ' << y << ' ' << z;
    
    return  0;
}

  

 

  
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总结

水题，不解释

文章来源: chen-ac.blog.csdn.net，作者：辰chen，版权归原作者所有，如需转载，请联系作者。

原文链接：chen-ac.blog.csdn.net/article/details/117912735