HDU 3571 N-dimensional Sphere( 高斯消元+ 同余 )

N-dimensional Sphere

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 668    Accepted Submission(s): 234

Problem Description

In an N-dimensional space, a sphere is defined as {(x1, x2 ... xN)| ∑(xi-Xi)^2 = R^2 (i=1,2,...,N) }. where (X1,X2…XN) is the center. You're given N + 1 points on an N-dimensional sphere and are asked to calculate the center of the sphere.

 

 

Input

The first line contains an integer T which is the number of test cases.
For each case there's one integer N on the first line.
Each of the N+1 following lines contains N integers x1, x2 ... xN describing the coordinate of a point on the N-dimensional sphere.
(0 <= T <= 10, 1 <= N <= 50, |xi| <= 10^17)

 

 

Output

For the kth case, first output a line contains “Case k:”, then output N integers on a line indicating the center of the N-dimensional sphere
(It's guaranteed that all coordinate components of the answer are integers and there is only one solution and |Xi| <= 10^17)

 

 

Sample Input

2

2

1 0

-1 0

0 1

3

2 2 3

0 2 3

1 3 3

1 2 4

 

 

Sample Output

Case 1:

0 0

Case 2:

1 2 3

 

 

 

这条题目的做法很容易想出来 。

凭借 n + 1 个点代入 n 维圆公式， 求圆心 。

然后用第 n + 1 个方程( 设下标为n )  sigma( ( Xi - Oi )^2 )  = R^2 

跟前n 个方程联立容易得到 ：

　　sigma( ( Xi - Oi )^2 )  =  sigma( ( Yi - Oi )^2 )  

两边都展开然后消掉Oi^2就得到

　　sigma(  2*( Xi - Yi )*Oi ) = sigma(  Xi^2 - Yi^2 )  .

得到 n 个这样的 n 元一次方程之后就可以利用高斯消元解决。

 

但首先 fabs( xi ) <= 1e17 的。 大数据的话显然计算过程溢出 。

就用到  sigma( ai * xi ) = an ( % mod ) 来解决。 求得解依然唯一。

 

在高斯消元的过程中会有除法 ， 用求逆来解决。

由于数据很大， 欧拉定理会溢 ， 那么用扩展欧几里得就OK 。

 

然后还需要将数据加一个偏移差，把所有数据处理成正数 （相当于把整个图形平移了，最后减回来不影响结果）。

避免在取余过程中把(负数+mod)%mod弄成了正。

 

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <algorithm>
using namespace std;
#define root 1,n,1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define lr rt<<1
#define rr rt<<1|1
typedef long long LL;
typedef pair<int,int>pii;
#define X first
#define Y second
const int oo = 1e9+7;
const double PI = acos(-1.0);
const double eps = 1e-6 ;
const int N = 55;
#define mod 200000000000000003LL
#define dif 100000000000000000LL

LL Mod(LL x) {
    if (x >= mod) return x - mod;
    return x;
}
LL mul(LL a, LL b) {
    LL res;
    for (res = 0; b; b >>= 1) {
        if (b & 1)
            res = Mod(res + a);
        a = Mod(a + a);
    }
    return res;
}

void e_gcd( LL a , LL b , LL &d , LL &x , LL &y ) {
    if( !b ){ d = a , x = 1 , y = 0 ; return ; }
    e_gcd( b , a%b , d , y , x );
    y -= x*(a/b);
}

LL inv( LL a , LL n ){
    LL d,x,y ;
    e_gcd(a,n,d,x,y);
    return ( x % n + n ) % n ;
}

LL A[N][N] , g[N][N];
int n ;

void Gauss() {

    for( int i = 0 ; i < n ; ++i ) {
        int r = i ;
        for( int j = i ; j < n ; ++j ) {
            if( g[j][i] ) { r = j ; break ; }
        }
        if( r != i ) for( int j = 0 ; j <= n ; ++j ) swap( g[i][j] , g[r][j] ) ;

        LL INV = inv( g[i][i] , mod );
        for( int k = i + 1 ; k < n ; ++k ) {
            if( g[k][i] ) {
                LL f = mul( g[k][i] , INV );
                for( int j = i ; j <= n ; ++j ) {
                    g[k][j] -= mul( f , g[i][j] );
                    g[k][j] = ( g[k][j] % mod + mod ) % mod ;
                }
            }
        }
    }
    for( int i = n - 1 ; i >= 0 ; --i ){
        for( int j = i + 1 ; j < n ; ++j ){
            g[i][n] -= mul( g[j][n] , g[i][j] ) , g[i][n] += mod , g[i][n] %= mod ;
        }
        g[i][n] = mul( g[i][n] , inv( g[i][i] , mod ) );
    }
}

void Run() {

    scanf("%d",&n);
    memset( g , 0 , sizeof g );
    for( int i = 0 ; i <= n ; ++i ) {
        for( int j = 0 ; j < n ; ++j ) {
            scanf("%I64d",&A[i][j]);
            A[i][j] += dif ;
        }
    }

    for( int i = 0 ; i < n ; ++i ){
        for( int j = 0 ; j < n ; ++j ){
            g[i][j] = Mod( A[n][j] - A[i][j] + mod );
            g[i][j] = mul( g[i][j] , 2 ) ;
            g[i][n] = Mod( g[i][n] + mul( A[n][j] , A[n][j] ) );
            g[i][n] = Mod( g[i][n] - mul( A[i][j] , A[i][j] ) + mod );
        }
    }

    Gauss();
    printf("%I64d",g[0][n]-dif);
    for( int i = 1 ; i < n ; ++i ){
        printf(" %I64d",g[i][n]-dif);
    }puts("");
}

int main()
{
    #ifdef LOCAL
        freopen("in.txt","r",stdin);
    #endif // LOCAL
    int cas = 1 , _ ; scanf("%d",&_ );
    while( _-- ){
        printf("Case %d:\n",cas++); Run();
    }
}