杭电 HDU 1570 A C

A C

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4206    Accepted Submission(s): 2671

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

 

Author

linle

 

Source

HDU 2007-1 Programming Contest

 

 哈尔滨生活也是醉了 ，放个假无聊到 刷这题儿玩儿。

#include<iostream>
using namespace std;
int fac(int k)
{
	int sum=1;
	for(int i=1;i<=k;i++)
		sum*=i;
	return sum;
}

int main()
{
	int n,N,M;
	char ch;
	cin>>n;
	while(n--)
	{
		cin>>ch>>N>>M;
		if(ch=='A')
			cout<<fac(N)/(fac(N-M))<<endl;
		else if(ch=='C')
			cout<<fac(N)/(fac(M)*fac(N-M))<<endl;
	}
	return 0;
}