#### What does "-nan" mean and what causes it?

I know that nan means "not a number",but this code using the Gaussian elimination algorithm output "-nan" :

``````#include <bits/stdc++.h>
#include <algorithm>

using namespace std;
typedef long long ll;
const int N=55;
const double eps=1e-7;
int n;

struct mat {
int l,c;
double p[N][N];
double* operator [] (int i) {return p[i];}
}a;

bool flo0(double &x) {return (fabs(x)<=eps)?1:0;}

void REF() {
for(int i=1;i<=n;i++) {
int flag=i;
for(int j=i;j<=n;j++) {
if(fabs(a[j][i])>fabs(a[flag][i])) flag=j;
}
for(int j=1;j<=n+1;j++) swap(a[flag][j],a[i][j]);
if(flo0(a[i][i])) continue;
for(int j=i+1;j<=n;j++) {
double t=a[j][i]/a[i][i];
for(int k=i;k<=n+1;k++) {
a[j][k]-=t*a[i][k];
}
}
}
}

void RREF() {
for(int i=n;i>=1;i--) {
if(flo0(a[i][i])) continue;
for(int j=n+1;j>=i;j--) a[i][j]/=a[i][i];
for(int j=i-1;j>=1;j--) {
double t=a[j][i];
for(int k=n+1;k>=1;k--) {
a[j][k]-=t*a[i][k];
}
}
}
bool no=false,many=false;
for(int i=1;i<=n;i++) {
int t=0;
for(int j=1;j<=n;j++) if(flo0(a[i][j])==false) ++t;
if(t==0 && flo0(a[i][n+1])==false) no=true;
if(t==0 && flo0(a[i][n+1])) many=true;
}
if(no) printf("-1");
else if(many) printf("0");
else for(int i=1;i<=n;i++) {if(flo0(a[i][n+1])) a[i][n+1]=0.0;printf("x%d=%.2lfn",i,a[i][n+1]);}
}

int main() {
scanf("%d",&n);
for(int i=1;i<=n;i++) {
for(int j=1;j<=n+1;j++) {
cin>>a[i][j];
}
}
REF();
RREF();

return 0;
}
``````

I can’t get the input,but the output is

``````x1=-nan
x2=-nan
x3=-nan
x4=-nan
x5=-nan
x6=-nan
x7=-nan
x8=-nan
x9=-nan
x10=-nan
x11=-nan
x12=-nan
x13=-nan
x14=-nan
x15=-nan
x1...
``````

I initially thought the reason is that I didn’t use "eps" when I determining if a double equals to 0,but after I added,the error still existed.