HDU 1570 A C （数学题）

A C

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4786    Accepted Submission(s): 3077

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 

AC代码：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<iomanip>
#include<algorithm>
#include<time.h>
typedef long long LL;
using namespace std;

int main()
{
	int a;
	cin>>a;
	while(a--)
	{
		char x;
		int m(0),n(0),sum;
		cin>>x;
		cin>>n>>m;
		sum=1;
		for(int i=n;i>n-m;i--)
		{
			sum=sum*i;
		}
	    if(x=='A') 
		cout<<sum<<endl;
		if(x=='C') 
		{
			int num=1;
			for(int j=m;j>0;j--)
			{
				num=num*j;
			}
			cout<<sum/num<<endl;
		}
	}

	return 0; 
}