CF552-Div3-A. Restoring Three Numbers

解密三数之谜

最新推荐文章于 2025-09-07 20:32:33 发布

转载最新推荐文章于 2025-09-07 20:32:33 发布 · 194 阅读

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原文链接：http://www.cnblogs.com/-yjun/p/10727216.html

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#数据结构与算法

本文介绍了一种解决数学谜题的方法，通过已知的四个数值（a+b, a+c, b+c, a+b+c），求解出三个未知正整数a、b、c的具体值。文章提供了一个简洁的C++代码实现，展示了如何通过排序和简单的数学运算来快速解决问题。

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).

题意:给你三个数,这三个数分别是a+b,a+c,b+c,a+b+c(无序);求出a,b,c.(a>0,b>0,c>0)

题解:存进数组,sort以后,输出x4-x1,x4-x2,x4-x3(any order).x4是四个数中最大的数,可得出x4=a+b+c.那么其他三个数就是a+b,a+c,b+c.

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
    int a[5];
    for(int i=1;i<=4;i++) cin>>a[i];
    sort(a+1,a+1+4);
    printf("%d %d %d\n",a[4]-a[1],a[4]-a[2],a[4]-a[3]);
    return 0;
}

转载于:https://www.cnblogs.com/-yjun/p/10727216.html