"Turn" hdu-2048-God, God, and heaven: Wrong row

Error-Line Simplification formula:

D (n) = [n!/e+0.5] where E is the base of the natural logarithm, [x] is the integer part of X

#include <iostream> #include <cmath> #include <iomanip> #include <cstdio>using namespace std; Const double E=EXP (1);//E 1-Time Square int main () {    double arr[21], tmp=1.0;    int T, n, I;    for (int i=2; i<21; i++)    {            tmp*=i;            Arr[i]=round (tmp/e)/tmp;//round function returns rounding integer value    }    scanf ("%d", &t);    while (t--)    {        scanf ("%d", &n);        printf ("%.2f%%\n", arr[n]*100.0);    } return 0;}

All possible permutations of the n note are naturally n! (denominator).
Now the problem is to ask for N-note the number of rows f (n) (molecule).
First of all we consider that if the front N-1 people are not their own votes, that is, the former N-1 individual satisfied the wrong row, now another person, he is holding his own ticket. As long as he exchanged his ticket with any of the other N-1 individuals, he would be able to satisfy the N-man's Wrong row. At this time there are (N-1) *f (N-1) methods.

#include <iostream> #include <string> #include <iomanip>using namespace Std;int main () {    double arr[21]={0, 0, 1};    for (int i=3; i<21; i++) {        arr[i]= (arr[i-1]+arr[i-2]) * (i-1);    }    int t;    cin>>t;    while (t--)    {            int n;            cin>>n;            Double tmp=1;            for (int i=2; i<=n; i++)                tmp*=i;            Cout<<fixed<<setprecision (2) << (arr[n]/tmp) *100.0<< "%" <<endl;    } return 0;}

　　

"Turn" hdu-2048-God, God, and heaven: Wrong row