codeforces723AThe New Year: Meeting Friends+水题

最新推荐文章于 2018-08-02 20:55:11 发布

xtulollipop

最新推荐文章于 2018-08-02 20:55:11 发布

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分类专栏：水题文章标签： codeforces

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本文介绍了一个简单的数学问题，即如何找到一条直线上三个不同点的最佳会合位置，使得从这三个点到达该位置的总距离最短。通过直接枚举的方法，我们可以轻松找到这个问题的答案。

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There are three friend living on the straight line Ox in Lineland. The first friend lives at the point x1, the second friend lives at the point x2, and the third friend lives at the point x3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?

It’s guaranteed that the optimal answer is always integer.
Input

The first line of the input contains three distinct integers x1, x2 and x3 (1 ≤ x1, x2, x3 ≤ 100) — the coordinates of the houses of the first, the second and the third friends respectively.
Output

Print one integer — the minimum total distance the friends need to travel in order to meet together.
Examples
Input

7 1 4

Output

Input

30 20 10

Output

Note

In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.

水题：给出x轴上3个点，求一个点到3个点距离最小。数据很小，直接暴力枚举就好了。

#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<queue>
#include<vector>
#include<map>
#include<stack>
#include<set>
#define pi acos(-1.0)
#define EPS 1e-6    //log(x)
#define e exp(1.0); //2.718281828
#define mod 1000000007
#define INF 0x7fffffff
#define inf 0x3f3f3f3f
typedef long long LL;
using namespace std;
int main()
{
    int x[5];
    scanf("%d %d %d",&x[0],&x[1],&x[2]);
    sort(x,x+3);
    int ans=inf;
    for(int i=x[0];i<=x[2];i++){
        ans=min(ans,abs(i-x[0])+abs(i-x[1])+abs(i-x[2]));
    }
    cout<<ans<<endl;
    return 0;
}