luogu 3389 【模板】高斯消元

大概就是对每一行先找到最大的减小误差，然后代入消元

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 #include<algorithm>
 #include<queue>
 #include<vector>
 #include<map>
 #include<set>
 #define ll long long
 #define db double
 #define inf 2139062143
 #define MAXN 200100
 #define rep(i,s,t) for(register int i=(s),i##__end=(t);i<=i##__end;++i)
 #define dwn(i,s,t) for(register int i=(s),i##__end=(t);i>=i##__end;--i)
 #define ren for(register int i=fst[x];i;i=nxt[i])
 #define Fill(x,t) memset(x,t,sizeof(x))
 #define pls(a,b) (a+b)%MOD
 #define mns(a,b) (a-b+MOD)%MOD
 #define mul(a,b) (1LL*(a)*(b))%MOD
 using namespace std;
 inline int read()
 {
     int x=,f=;char ch=getchar();
     while(!isdigit(ch)) {if(ch=='-') f=-;ch=getchar();}
     while(isdigit(ch)) {x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 int n;db a[][];
 void Gauss()
 {
     int pos;rep(i,,n-)
     {
         pos=n;rep(j,i,n-)if(fabs(a[j][i])>fabs(a[pos][i])) pos=j;
         if(pos==n) {puts("No Solution");exit();}
         if(pos!=i) rep(j,,n) swap(a[i][j],a[pos][j]);
         dwn(j,n,i) rep(k,i+,n-) a[k][j]-=a[k][i]/a[i][i]*a[i][j];
     }
     dwn(i,n-,) {rep(j,i+,n-) a[i][n]-=a[j][n]*a[i][j];a[i][n]/=a[i][i];}
 }
 int main()
 {
     n=read();rep(i,,n-) rep(j,,n) a[i][j]=read();Gauss();
     rep(i,,n-) printf("%.2lf\n",a[i][n]);
 }