坑。。待填

 http://acm.hdu.edu.cn/showproblem.php?pid=1031

http://www.cnblogs.com/chanme/p/3861766.html

http://m.blog.csdn.net/blog/tjdrn/9329531

http://blog.csdn.net/xuezhongfenfei/article/details/9822173

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2619

hdu 3571 N-dimensional Sphere 高斯消元

#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 //需要的是素数 
#define dif 100000000000000000LL //偏移量，使得数都是整数，方便移位乘法 

LL Mod(LL x) { //加法取模，防止超__int64  
    if (x >= mod) return x - mod;
    return x;
}
LL mul(LL a, LL b) { //乘法取模，用移位乘法，防止超__int64 
    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 ) { //拓展的欧几里德定理，求ax+by=gcd(a,b)的一个解 
    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 ){ //求逆，用于除法取模 
    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() { //高斯消元 

    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 );//相当于g[k][i]/g[i][i]%mod;  
                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 ; //偏移diff  
        }
    }

    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); //减去先前偏移的值
    for( int i = 1 ; i < n ; ++i ){
        printf(" %I64d",g[i][n]-dif);
    }puts("");
}

int main()
{
    int cas = 1 , _ ; scanf("%d",&_ );
    while( _-- ){
        printf("Case %d:\n",cas++); Run();
    }
}

 //扩展欧几里德算法
 int ExGCD(int a, int b, int& x, int& y)
 {
     if(b == 0)
     {
         x = 1, y = 0;
         return a;
     }
     int d = ExGCD(b, a%b, x, y);
     int temp = x;
     x = y;
     y = temp - a/b*y;
     return d;
 }

 int main()
 {
     int x, y, d;
     d = ExGCD(99, 78, x, y);
     cout << d << " " << x << " " << y << endl;
     return 0;
 }

 //定理一: 如果a,b是不都为0的任意整数,则d=gcd(a,b)是a,b的线性组合{ax+by: x,y∈Z}的最小元素.
 // 已知d=gcd(a,b)=gcd(b,a mod b)
 //
 //由gcd(b,a mod b)得知,d = bx + a mod b = bx + (a-floor(a/b)*b)*y = a*y + b(x-floor(a/b)*y)
 //当推到gcd(a,b)时,d′ = d = a*y + b(x-floor(a/b)*y)

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