AC

A C

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1741    Accepted Submission(s): 1089

Problem Description

Are you excited when you see the title "AC" ? If the answer is YES , AC it ;

You must learn these two combination formulas in the school . If you have forgotten it , see the picture.




Now I will give you n and m , and your task is to calculate the answer .

 

Input

In the first line , there is a integer T indicates the number of test cases.
Then T cases follows in the T lines.
Each case contains a character 'A' or 'C', two integers represent n and m. (1<=n,m<=10)

 

Output

For each case , if the character is 'A' , calculate A(m,n),and if the character is 'C' , calculate C(m,n).
And print the answer in a single line.

 

Sample Input

2 A 10 10 C 4 2

 

Sample Output

3628800 6

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int c[20];
int a[40];
char str[10];

int pow( int x )
{   int i, sum = 1;
    for (i = 1; i <= x; i++)
        sum *= i;
    return sum;
}

int fun( int n, int v)
{
     int i = 0;
     c[0] = 1;
     for (i = 1; i <= n; i++) {
         c[i] = c[i-1] * (n - i + 1 ) / i;
     }
      if ( str[0] == 'C')
           return c[v];
      else
           return c[v] * pow(v);
}
         
int main( )
{
    int T, a, b;
    scanf("%d", &T);
    while (T--)
    {
          scanf("%s%d%d",str, &a, &b);
          printf("%d\n",fun(a,b));
    }
    return 0;
}