A C（链接）
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 .
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)
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.
And print the answer in a single line.
Sample Input
2 A 10 10 C 4 2
Sample Output
3628800 6
因为n,m的范围比较小，直接求出 1 - 10 的阶乘就好了
#include<cstdio>
#include<algorithm>
using namespace std;int T,n,m,a[20];
char c;//计算阶乘 注意 0
void init()
{int ans=1;a[0]=1;for(int i=1;i<11;i++){a[i]=ans*i;ans=a[i];}
}int A(int n,int m)
{return a[n]/a[n-m];
}int C(int n,int m)
{return a[n]/a[m]/a[n-m];
}int main()
{int num;init();scanf("%d",&T);while(T--){getchar(); //这里接收回车scanf("%c %d %d",&c,&n,&m);if(c == 'A')num = A(n,m);else if(c == 'C')num = C(n,m);printf("%d\n",num);}return 0;
}
