HDU 1570 A C

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本文介绍了一个简单的AC组合计算问题，通过给定的n和m值，根据字符指示计算A(m,n)或C(m,n)，并提供了完整的C语言实现代码。

/*
解题人：lingnichong
解题时间：2014-09-02 11:39:05
解题体会：简单题
*/

A C

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3803 Accepted Submission(s): 2432

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

Sample Output

  
   3628800
6

Author

linle

Source

HDU 2007-1 Programming Contest

#include<stdio.h>
int main()
{
    int t;
    char a;
    int m,n,i;
    scanf("%d",&t);
    getchar();
    while(t--)
    {
        int sum=1,sum1=1;
        scanf("%c%d%d",&a,&n,&m);
        getchar();
        if(a=='A')
        {
            for(i=n-m+1;i<=n;i++)
            sum*=i;
        } 
        else if(a=='C') 
        {
            if(m<n-m)
           		 m=n-m;
 	         for(i=1;i<=m;i++)
       		 	sum1*=i;
             for(i=n-m+1;i<=n;i++)
             	sum*=i;
         	 sum=sum/sum1;
        }       
        printf("%d\n",sum);
    }
    return 0;
}