hdoj5626Clarke and points

Clarke and points

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 107    Accepted Submission(s): 81

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

2
3 233
5 332

 

Sample Output

1557439953
1423870062

 

//xa-xb+ya-yb
//xa+ya-(xb+yb);
//xa-xb+yb-ya
//xa-ya+yb-xb;
//
//xb-xa+ya-yb
//ya-xa+xb-yb;
//xb-xa+yb-ya 
//xb+yb-(xa+ya);

#include<cstdio>
#include<cstring>
#include<algorithm>
#define inf 50000000000000
using namespace std;
long long n,seed,x[1000010],y[1000010];
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()
{
    long long t;
    scanf("%I64d",&t);
    while(t--){    
    long long x1=inf,y1=inf,x2=-inf,y2=-inf,ans1=-inf,ans2=inf,ans3=-inf,ans4=-inf;
    scanf("%I64d%I64d",&n,&seed);
    for (int i = 0; i < n; i++){
        x[i] = rand(-1000000000, 1000000000);
        y[i] = rand(-1000000000, 1000000000);
        ans1=max(x[i]+y[i],ans1);ans2=min(x[i]+y[i],ans2);
        ans3=max(x[i]-y[i],ans3);ans4=max(ans4,y[i]-x[i]);
    }
    printf("%I64d\n",max(ans1-ans2,ans3+ans4));
    }
    return 0;
}