hdu 5626 Clarke and points

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

Source

BestCoder Round #72 (div.2)

先附上自己的写法，运气好的话可以过，运气不好的话超时，这东西也看人品？

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<math.h>
 #include<algorithm>
 #include<queue>
 #include<set>
 #include<bitset>
 #include<map>
 #include<vector>
 #include<stdlib.h>
 using namespace std;
 #define ll long long
 #define eps 1e-10
 #define MOD 1000000007
 #define N 1000006
 #define inf 1e12

 struct node{
     ll x,y;
 }e[N],res[N];
 ll cmp(node a,node b)
 {
     if(a.x==b.x)return a.y<b.y;
     return a.x<b.x;
 }
 ll cross(node a,node b,node c)//向量积
 {
     return (a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y);
 }
 ll convex(ll n)//求凸包上的点
 {
     sort(e,e+n,cmp);
     ll m=,i,j,k;
     //求得下凸包，逆时针
     //已知凸包点m个，如果新加入点为i，则向量(m-2,i)必定要在(m-2,m-1)的逆时针方向才符合凸包的性质
     //若不成立，则m-1点不在凸包上。
     for(i=;i<n;i++)
     {
         while(m>&&cross(res[m-],e[i],res[m-])<=)m--;
         res[m++]=e[i];
     }
     k=m;
     //求得上凸包
     for(i=n-;i>=;i--)
     {
         while(m>k&&cross(res[m-],e[i],res[m-])<=)m--;
         res[m++]=e[i];
     }
     if(n>)m--;//起始点重复。
     return m;
 }  

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

 int main()
 {
     int t;
     scanf("%d",&t);
     while(t--){
         cin >> n >> seed;
         for (int i = ; i < n; i++){
             e[i].x = rand(-, ),
             e[i].y = rand(-, );
         }
         ll    m=convex(n);
         ll ans=-;
         for(ll i=;i<m;i++){
             for(ll j=i+;j<m;j++){
                 ll cnt = abs(res[i].x-res[j].x)+abs(res[i].y-res[j].y);
                 ans=max(ans,cnt);
             }
         }
         printf("%I64d\n",ans);

     }
     return ;
 }

官方题解：

 #include<bitset>
 #include<map>
 #include<vector>
 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<string>
 #include<algorithm>
 #include<cmath>
 #include<stack>
 #include<queue>
 #include<set>
 #define inf 0x3f3f3f3f
 #define mem(a,x) memset(a,x,sizeof(a))

 using namespace std;

 typedef long long ll;
 typedef unsigned long long ull;
 typedef pair<int,int> pii;

 inline int in()
 {
     int res=;char c;int f=;
     while((c=getchar())<'' || c>'')if(c=='-')f=-;
     while(c>='' && c<='')res=res*+c-'',c=getchar();
     return res*f;
 }
 const int  N = ;

 ll a[N][];
 int n;
 long long seed;
 inline long long rand(long long l, long long r) {
     static long long mo=1e9+, g=;
     return l+((seed*=g)%=mo)%(r-l+);
 }
 int main() {
     int T;
     for (scanf("%d", &T);T--;) {
         cin >> n >> seed;
         for (int i=; i<n; i++)
             a[i][]=rand(-, ),
             a[i][]=rand(-, );
         ll t=;
         ll ans=,mx=-9223372036854775808LL,mn=9223372036854775807LL;
         for (int s=; s<(<<); s++) {
             mx=-9223372036854775808LL,mn=9223372036854775807LL;
             for (int i=; i<n; i++) {
                 t = ;
                 for (int j=; j<; j++)
                     if ((<<j) & s) t += a[i][j];
                     else t -= a[i][j];
                 mn = min(mn, t);
                 mx = max(mx, t);
             }
             ans = max(ans, mx-mn);
         }
         printf("%I64d\n", ans);
     }
     return ;
 }