Cod sursa(job #1341874)

Utilizator Kira96Denis Mita Kira96 Data 13 februarie 2015 10:47:10
Problema Algoritmul lui Gauss Scor 80
Compilator cpp Status done
Runda Arhiva educationala Marime 2.09 kb
/*
    Look at me!
    Look at me!
    Look at how large the monster inside me has become!
*/

#include<fstream>
#include<cstdio>
#include<map>
#include<set>
#define FIT(a,b) for(vector<int >::iterator a=b.begin();a!=b.end();a++)
#define FITP(a,b) for(vector<pair<int,int> >::iterator a=b.begin();a!=b.end();a++)
#define RIT(a,b) for(vector<int>::reverse_iterator a=b.end();a!=b.begin();++a)
#include<stack>
#define ROF(a,b,c) for(int a=b;a>=c;--a)
#include<vector>
#include<algorithm>
#define FOR(a,b,c) for(int a=b;a<=c;++a)
#define REP(a,b) for(register int a=0;a<b;++a)
#include<cstring>
#include<ctime>
#include<bitset>
#include<cmath>
#include<iomanip>
#include<set>
#define f cin
#define g cout
#include<queue>
#define debug cerr<<"OK";
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ll long long
#define ull unsigned long long
#define mod 1000000007
#define MOD 32416190071
#define N 1010
#define SQR 350
#define inf 1<<30
#define eps 1.e-4
using namespace std;

ifstream f("gauss.in");
ofstream g("gauss.out");

double X[N],A[N][N];
int n,m;
int zero(double x)
{
    if(x<eps&&x>-eps)
    return 1;
    return 0;
}
int main ()
{
    f>>n>>m;
    FOR(i,1,n)
    FOR(j,1,m+1)
    f>>A[i][j];

    FOR(j,1,m)
    {
        int p=n+1;
        FOR(i,j,n)
        if(!zero(A[i][j]))
        {
            p=i;
            break;
        }
        if(p==n+1)
        continue;
        if(p!=j)
        FOR(i,1,m+1)
        swap(A[j][i],A[p][i]);
        FOR(i,j+1,m+1)
        A[j][i]/=A[j][j];
        A[j][j]=1;

        FOR(k,j+1,n)
        {
            FOR(i,j+1,m+1)
            A[k][i]-=A[k][j]*A[j][i];
            A[k][j]=0;
        }
    }
    ROF(i,n,1)
    FOR(j,1,m+1)
    if(!zero(A[i][j]))
    {
        if(j==m+1)
        {
            g<<"Imposibil";
            return 0;
        }
        X[j]=A[i][m+1];
        FOR(k,j+1,m)
        X[j]-=X[k]*A[i][k];
        break;
    }
    FOR(i,1,m)
    g<<fixed<<setprecision(11)<<X[i]<<" ";
    return 0;
}