Clarke and points

题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=5626

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

Hint

题意

让你求平面两点的曼哈顿最远距离

题解：

显然我们可以看出距离 = abs(x1-x2)+abs(y1-y2)

我们把绝对值拆开，然后再归纳一下，显然可以分为一下四种情况(x1+y1)-(x2+y2),(x1-y1)-(x2-y2),(-x1+y1)-(-x2+y2),(-x1-y1)-(-x2-y2)

我们可以看出减号左右是相同的，所以我们维护这四个值的最大最小值就好了

代码

#include<algorithm>
#include<iostream>
#include<vector>
using namespace std;
const int maxn = 1e6+7;
int n;
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);
}
long long Max[10];
long long Min[10];
int main()
{
    int t;
    scanf("%d",&t);
    for(int cas=1;cas<=t;cas++)
    {
        cin >> n >> seed;
        for(int i=0;i<10;i++)
            Max[i]=-1e15,Min[i]=1e15;
        long long x,y;
        for (int i = 0; i < n; i++)
        {
            x = rand(-1000000000, 1000000000),
            y = rand(-1000000000, 1000000000);
            Max[0]=max(Max[0],x+y);
            Max[1]=max(Max[1],-x+y);
            Max[2]=max(Max[2],x-y);
            Max[3]=max(Max[3],-x-y);
            Min[0]=min(Min[0],x+y);
            Min[1]=min(Min[1],-x+y);
            Min[2]=min(Min[2],x-y);
            Min[3]=min(Min[3],-x-y);
        }
        long long ans = 0;
        for(int i=0;i<4;i++)
            ans=max(Max[i]-Min[i],ans);
        cout<<ans<<endl;
    }
}