You can not select more than 25 topics
Topics must start with a letter or number, can include dashes ('-') and can be up to 35 characters long.
|
|
|
|
#include <bits/stdc++.h>
|
|
|
|
|
using namespace std;
|
|
|
|
|
|
|
|
|
|
const int N = 110;
|
|
|
|
|
const double eps = 1e-8;
|
|
|
|
|
|
|
|
|
|
int n;
|
|
|
|
|
double a[N][N];
|
|
|
|
|
|
|
|
|
|
/*
|
|
|
|
|
高斯消元,答案存于a[i][n]中,0 <= i < n
|
|
|
|
|
0:唯一解 1:无穷多组解 2:无解
|
|
|
|
|
*/
|
|
|
|
|
int gauss() {
|
|
|
|
|
int c, r;
|
|
|
|
|
for (c = 0, r = 0; c < n; c++) { //枚举的是列
|
|
|
|
|
int t = r;
|
|
|
|
|
for (int i = r; i < n; i++) // 找绝对值最大的行,可以解决精度问题
|
|
|
|
|
if (abs(a[i][c]) > abs(a[t][c])) t = i;
|
|
|
|
|
|
|
|
|
|
if (abs(a[t][c]) < eps) continue; //是0不处理
|
|
|
|
|
|
|
|
|
|
swap(a[t], a[r]); // 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,倒着来
|
|
|
|
|
for (int i = r + 1; i < n; i++) // 用当前行将下面所有的列消成0
|
|
|
|
|
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 (abs(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() {
|
|
|
|
|
cin >> n;
|
|
|
|
|
for (int i = 0; i < n; i++)
|
|
|
|
|
for (int j = 0; j < n + 1; j++) //带等式右侧的结果值
|
|
|
|
|
cin >> a[i][j];
|
|
|
|
|
|
|
|
|
|
int t = gauss();
|
|
|
|
|
if (t == 2) //无解
|
|
|
|
|
puts("No solution");
|
|
|
|
|
else if (t == 1) //无穷多组
|
|
|
|
|
puts("Infinite group solutions");
|
|
|
|
|
else {
|
|
|
|
|
//唯一解
|
|
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
|
|
if (abs(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;
|
|
|
|
|
}
|