【luogu P2455 [SDOI2006]线性方程组】 题解

题目链接：https://www.luogu.org/problemnew/show/P2455

无解：最后一列对应元素不为0，前面全是0.

无穷解：一行全是0.

嗯...在消元过程中不要直接拿矩阵元素自己消，会把自己消成0.

#include <algorithm>
#include <cstdio>
#include <cmath>
#include <iostream>
using namespace std;
const int maxn = 200;
const double eps = 1e-7;
double A[maxn][maxn], ans[maxn];
int n;
int main() 
{
    scanf("%d",&n);
    for(int i = 1; i <= n; i++)
    for(int j = 1; j <= n+1; j++)
    scanf("%lf", &A[i][j]);
    for(int i = 1; i <= n; i++) 
    {
        int p = i;
        for(int j = i + 1; j <= n; j++)
          if(fabs(A[j][i]) > fabs(A[p][i])) p = j;
        for(int j = 1; j <= n + 1; j++) swap(A[p][j],A[i][j]);
          
        if(fabs(A[i][i]) < eps) continue;
        double div = A[i][i];
        for(int j = 1; j <= n + 1; j++) A[i][j]/=div;
        for(int j = 1; j <= n; j++)
        {
            if(i != j)
            {
                double div = A[j][i];
                for(int k = 1; k <= n + 1; k++) A[j][k] -= A[i][k]*div;
            }
        }
    }
    int NoSolution = 0, ManySolution = 0;
    for(int i = 1; i <= n; i++)
    {
        int Nonum = 0, Manynum = 0;
        for(int j = 1; j <= n + 1 && fabs(A[i][j]) < eps; j++)
        Nonum++,Manynum++;
        if(Manynum > n) ManySolution = 1;
        if(Nonum == n) NoSolution = 1;
    }
        if(NoSolution) {printf("-1");return 0;}
        if(ManySolution) {printf("0");return 0;}
    for(int i = n; i >= 1; i--)
    {
        ans[i] = A[i][n+1];
        for(int j = i - 1; j >= 1; j--)
        {
            A[j][n+1] -= ans[i] * A[j][i];
            A[j][i] = 0;
        }
    }
    for(int i = 1; i <= n; i++)
    printf("x%d=%.2lf
",i,ans[i] + eps);
    return 0;
}