hdoj 1570 A C

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 .

 

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

2A 10 10C 4 2

 

Sample Output

36288006

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 0void 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;}