Luogu2455 [SDOI2006]线性方程组 （高斯消元）

模板特殊情况没exit(0) $\longrightarrow$60 了一下午

//#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int  a = (b); a <= (c); ++ a)
#define nR(a,b,c) for(register int  a = (b); a >= (c); -- a)
#define Max(a,b) ((a) > (b) ? (a) : (b))
#define Min(a,b) ((a) < (b) ? (a) : (b))
#define Fill(a,b) memset(a, b, sizeof(a))
#define Abs(a) ((a) < 0 ? -(a) : (a))
#define Swap(a,b) a^=b^=a^=b
#define ll long long

#define ON_DEBUG

#ifdef ON_DEBUG

#define D_e_Line printf("\n\n----------\n\n")
#define D_e(x)  cout << #x << " = " << x << endl
#define Pause() system("pause")
#define FileOpen() freopen("in.txt","r",stdin);

#else

#define D_e_Line ;
#define D_e(x)  ;
#define Pause() ;
#define FileOpen() ;

#endif

struct ios{
    template<typename ATP>ios& operator >> (ATP &x){
        x = 0; int f = 1; char c;
        for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-')  f = -1;
        while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
        x*= f;
        return *this;
    }
}io;
using namespace std;

const int N = 107;
const double eps = 1e-8;

int n;
double a[N][N], ans[N];
inline double fabs(double a){
	return a < 0 ? -a : a;
}
inline void Gauss(){
	R(i,1,n){
		int r = i;
		R(j,i + 1,n)
			if(fabs(a[j][i]) > fabs(a[r][i]))
				r = j;
		if(i != r) swap(a[i], a[r]);
		if(fabs(a[i][i]) > eps){
			R(j,1,n){
				if(i == j) continue;
				double s = a[j][i] / a[i][i];
				R(k,1,n + 1)
					a[j][k] -= s * a[i][k];
			}
		}
	}
	int flagNo = 0, flagInf = 0;
	R(i,1,n){
		int tot = 0;
		R(j,1,n + 1){
			if(fabs(a[i][j]) < eps)
				++tot;
			else
				break;
		}
		if(tot == n + 1) flagInf = 1;
		else if(tot == n && fabs(a[i][n + 1]) > eps) flagNo = 1;
	}
	if(flagNo == 1){ printf("-1"); exit(0);}
	if(flagInf == 1){ printf("0"); exit(0);}

	nR(i,n,1){
		ans[i] = a[i][n + 1] / a[i][i];
		nR(j,i - 1,1){
			a[j][n + 1] -= a[j][i] * ans[i];
		}
	}
}

int main(){
	io >> n;
	R(i,1,n){
		R(j,1,n + 1){
			scanf("%lf", &a[i][j]);
		}
	}

	Gauss();

	R(i,1,n){
//		if(fabs(ans[i]) < eps)
//			printf("x%d=0\n", i);
//		else
			printf("x%d=%.2lf\n", i, ans[i]);
	}

	return 0;
}