高斯消元

#include<bits/stdc++.h>
#define re return
#define inc(i,l,r) for(int i=l;i<=r;++i)
#define dec(i,l,r) for(int i=l;i>=r;--i)
const double EPS=1e-7;

char buf[1<<21],*p1,*p2;
inline int getc(){re p1==p2 and (p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}

template<typename T>inline void rd(T&x)
{
    char c;bool f=0;
    while((c=getc())<'0'||c>'9')if(c=='-')f=1;
    x=c^48;
    while((c=getc())>='0' and c<='9')x=x*10+(c^48);
    if(f)x=-x;
}
using namespace std;
double a[105][105],ans[105];
int n;

int main()
{
//    freopen("in.txt","r",stdin);
    rd(n);
    inc(i,1,n)inc(j,1,n+1)rd(a[i][j]);


    inc(i,1,n){
        int r=i;
        if(fabs(a[r][r])<EPS){
            inc(j,i+1,n)
            if(a[j][r]){swap(a[i],a[j]);break;}
            if(fabs(a[i][i])<EPS){printf("No Solution"); re 0;}//存在自由元 
        }

        double div=a[i][i];
        inc(j,i,n+1)a[i][j]/=div;//这一行系数化为1,1-i-1都已是0 
        inc(j,i+1,n){
            div=a[j][i];
            inc(k,i,n+1)a[j][k]-=a[i][k]*div;//弄去剩下行（i）系数
            //但可能一不小心弄多（把某一列的z系数也化没有了）了 
        }
    }
    dec(i,n,1){
        ans[i]=a[i][n+1];//得解 
        dec(j,i-1,1)a[j][n+1]-=a[j][i]*ans[i];//代回去 
    }

    inc(i,1,n)
    printf("%.2lf\n",ans[i]);
    re 0;
}

线性函数，加减消元