坑。。待填

来源：互联网发布：淘宝记账软件哪个好编辑：程序博客网时间：2024/05/19 16:35

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|1typedef long long LL;typedef pair<int,int>pii;#define X first#define Y secondconst 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)

http://wenku.baidu.com/link?url=bpQWehZg4Qn83SAylTTrgFTz7N0t2pEthJlGEO1X01_InkxfR-_UH0vrrah4VucfNvyEkMHhhOX9fG6zF9_12y_O38pYEOup9pQX-2yV7ty

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