Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P2455 [SDOI2006]线性方程组

P2455 [SDOI2006]线性方程组

高斯消元模板。

查看代码
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#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
const int N = 110;
const double eps = 1e-4;
int n;
double A[N][N];
int Gauss()
{
int c = 0, r = 0;
for (; c < n; c++)
{
int t = r;
for (int i = r; i < n; i++)
if (fabs(A[i][c]) > fabs(A[t][c]))
t = i;
if (fabs(A[t][c]) < eps)
continue;
for (int i = c; i <= n; i++)
swap(A[t][i], A[r][i]);
for (int i = n; i >= c; i--)
A[r][i] /= A[r][c];
for (int i = r + 1; i < n; i++)
if (fabs(A[i][c]) > eps)
for (int j = n; j >= c; j--)
A[i][j] -= A[r][j] * A[i][c];
r++;
}
if (r < n)
{
for (int i = r; i < n; i++)
if (fabs(A[i][n]) > eps)
return 1; // 无解
return 2; // 无穷多组解
}
for (int i = n - 1; i >= 0; i--)
for (int j = i + 1; j < n; ++j)
A[i][n] -= A[i][j] * A[j][n];
return 0;
}
int main()
{
cin >> n;
for (int i = 0; i < n; i++)
for (int j = 0; j <= n; j++)
cin >> A[i][j];
int t = Gauss();
if (t == 0)
for (int i = 0; i < n; i++)
printf("x%d=%.2f\n", i + 1, A[i][n]);
else if (t == 1)
cout << -1;
else if (t == 2)
cout << 0;
return 0;
}