HDU 5626 Clarke and points-优快云博客

本文介绍了一个关于计算最大曼哈顿距离的问题。患者Clarke患有多重人格障碍，在某一天转变成几何学者，研究如何从随机生成的大量点中找到两两之间的最大曼哈顿距离。文章详细阐述了问题描述、输入输出格式、样例输入输出以及解决方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Clarke and points

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 127 Accepted Submission(s): 98

Problem Description

Clarke is a patient with multiple personality disorder. One day he turned into a learner of geometric.
He did a research on a interesting distance called Manhattan Distance. The Manhattan Distance between point

A(xA,yA) and point

B(xB,yB) is

|xA−xB|+|yA−yB| .
Now he wants to find the maximum distance between two points of

n points.

Input

The first line contains a integer

T(1≤T≤5) , the number of test case.
For each test case, a line followed, contains two integers

n,seed(2≤n≤1000000,1≤seed≤109) , denotes the number of points and a random seed.
The coordinate of each point is generated by the followed code.

```
long long seed;
inline long long rand(long long l, long long r) {
  static long long mo=1e9+7, g=78125;
  return l+((seed*=g)%=mo)%(r-l+1);
}

// ...

cin >> n >> seed;
for (int i = 0; i < n; i++)
  x[i] = rand(-1000000000, 1000000000),
  y[i] = rand(-1000000000, 1000000000);
```

Output

For each test case, print a line with an integer represented the maximum distance.

Sample Input

Sample Output

  
   1557439953
1423870062

#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;

typedef long long ll;
const int maxn = 1000000 + 10;
struct point{
    ll x, y;
}a[maxn];

ll seed;
inline long long Rand(long long l,long long r) {
    static long long mo=1e9+7, g=78125;
    return l+((seed*=g)%=mo)%(r-l+1);
}

int main(){
    int t,n;
    cin>>t;
    while(t--){
        cin>>n>>seed;
        for(int i = 0; i < n; i++){
            a[i].x = Rand(-1000000000, 1000000000),
            a[i].y = Rand(-1000000000, 1000000000);
        }
        ll xMax = a[0].x + a[0].y, xMin = a[0].x + a[0].y;
        ll yMax = a[0].x - a[0].x, yMin = a[0].x - a[0].y;
        for(int i = 1; i < n; i++){
            ll ax = a[i].x + a[i].y;
            ll in = a[i].x - a[i].y;
            xMax = max(xMax, ax);
            xMin = min(xMin, ax);
            yMax = max(yMax, in);
            yMin = min(yMin, in);
        }
        ll ans = max(abs(xMax - xMin), abs(yMax - yMin));
        cout<<ans<<endl;
    }
    return 0;
}