题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1570

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.

题解报告:hdu 1570 A C-LMLPHP

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

解题思路:组合和全排列!(水题~)

AC代码:

 #include<bits/stdc++.h>
using namespace std;
int main()
{
int T,n,m,sum;
char ch;
cin>>T;
while(T--){
getchar();//吃掉回车符
cin>>ch>>n>>m;
if(m>n)swap(n,m);//保证m比n小
sum=;//计数
if(ch=='A')for(int i=m;i>=;i--)sum*=(n-i+);//全排列
if(ch=='C'){//组合
if(n-m<m)m=n-m;//取m最小,减少计算
for(int i=;i<=m;i++)sum=sum*(n-i+)/i;
}
cout<<sum<<endl;
}
return ;
}
05-22 01:59
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