# 4、高斯消元线性方程组
# AcWing 883. 高斯消元解线性方程组
# AcWing 884. 高斯消元解异或线性方程组
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int N = 110;
const double eps = 1e-8;
int n;
double a[N][N];
int gauss() // 高斯消元,答案存于a[i][n]中,0 <= i < n
{
int c, r;
for (c = 0, r = 0; c < n; c ++ )
{
int t = r;
for (int i = r; i < n; i ++ ) // 找绝对值最大的行
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]); // 将绝对值最大的行换到最顶端
for (int i = n; i >= c; i -- ) a[r][i] /= a[r][c]; // 将当前行的首位变成1
for (int i = r + 1; i < n; i ++ ) // 用当前行将下面所有的列消成0
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 2; // 无解
return 1; // 有无穷多组解
}
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; // 有唯一解
}
int main()
{
scanf("%d", &n);
for (int i = 0; i < n; i ++ )
for (int j = 0; j < n + 1; j ++ )
scanf("%lf", &a[i][j]);
int t = gauss();
if (t == 2) puts("No solution");
else if (t == 1) puts("Infinite group solutions");
else
{
for (int i = 0; i < n; i ++ )
{
if (fabs(a[i][n]) < eps) a[i][n] = 0; // 去掉输出 -0.00 的情况
printf("%.2lf\n", a[i][n]);
}
}
return 0;
}
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#include <iostream>
#include <algorithm>
using namespace std;
const int N = 110;
int n;
int a[N][N];
int gauss()
{
int c, r;
for (c = 0, r = 0; c < n; c ++ )
{
int t = r;
for (int i = r; i < n; i ++ )
if (a[i][c])
t = i;
if (!a[t][c]) continue;
for (int i = c; i <= n; i ++ ) swap(a[r][i], a[t][i]);
for (int i = r + 1; i < n; i ++ )
if (a[i][c])
for (int j = n; j >= c; j -- )
a[i][j] ^= a[r][j];
r ++ ;
}
if (r < n)
{
for (int i = r; i < n; i ++ )
if (a[i][n])
return 2;
return 1;
}
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 + 1; j ++ )
cin >> a[i][j];
int t = gauss();
if (t == 0)
{
for (int i = 0; i < n; i ++ ) cout << a[i][n] << endl;
}
else if (t == 1) puts("Multiple sets of solutions");
else puts("No solution");
return 0;
}
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