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hdu 1570 AC

时间:2016-06-05 13:58:10      阅读:225      评论:0      收藏:0      [点我收藏+]

A C

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4830    Accepted Submission(s): 3115

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.
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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
 1 #include <iostream>
 2 using namespace std;
 3 int f[11] = {1, 1};
 4 
 5 
 6 int main(){
 7     for(int i = 2; i < 11; i++){
 8         f[i] = i * f[i-1];
 9     }
10     int test, n, m;
11     char c;
12     cin >> test;
13     while(test--){
14         cin >> c >> n >> m;
15         if(c == A)
16             cout << f[n] / f[n - m] << endl;
17         else if(c == C)
18             cout << f[n] / f[m] / f[n - m] << endl;
19     }
20     return 0;
21 }

 

hdu 1570 AC

原文:http://www.cnblogs.com/qinduanyinghua/p/5560546.html

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