http://acm.hdu.edu.cn/showproblem.php?pid=1031
http://www.cnblogs.com/chanme/p/3861766.html
http://m.blog.youkuaiyun.com/blog/tjdrn/9329531
http://blog.youkuaiyun.com/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)