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  • Friends

    Description

    Mike has many friends. Here are nine of them: Alice, Bob, Carol, Dave, Eve, Frank, Gloria, Henry and Irene. 

    Mike is so skillful that he can master $n$ languages (aka. programming languages). 

    His nine friends are all weaker than he. The sets they can master are all subsets of Mike's languages. 

    But the relations between the nine friends is very complex. Here are some clues. 

    1. Alice is a nice girl, so her subset is a superset of Bob's. 
    2. Bob is a naughty boy, so his subset is a superset of Carol's. 
    3. Dave is a handsome boy, so his subset is a superset of Eve's. 
    4. Eve is an evil girl, so her subset is a superset of Frank's. 
    5. Gloria is a cute girl, so her subset is a superset of Henry's. 
    6. Henry is a tall boy, so his subset is a superset of Irene's. 
    7. Alice is a nice girl, so her subset is a superset of Eve's. 
    8. Eve is an evil girl, so her subset is a superset of Carol's. 
    9. Dave is a handsome boy, so his subset is a superset of Gloria's. 
    10. Gloria is a cute girl, so her subset is a superset of Frank's. 
    11. Gloria is a cute girl, so her subset is a superset of Bob's. 

    Now Mike wants to know, how many situations there might be. 
     

    Input

    The first line contains an integer $T$($T le 20$) denoting the number of test cases. 

    For each test case, the first line contains an integer $n$($0leq nleq 3000$), denoting the number of languages. 

     

    Output

    For each test case, output ''Case #t:'' to represent this is the t-th case. And then output the answer. 
     

    Sample Input

    2 0 2
     

    Sample Output

    Case #1: 1
    Case #2: 1024
     只要读懂题目就知道了
    其实就是求32的n次方; 大数问题用Java是很方便的
     1 import java.math.BigInteger;
     2 import java.util.Scanner;
     3 public class Main{
     4 
     5     public static void main(String[] args) {
     6         Scanner in=new Scanner(System.in);
     7         int t=in.nextInt();
     8         int cased=1;
     9         while(t-->0){
    10             int n=in.nextInt();
    11             BigInteger b=new BigInteger("32");
    12             System.out.println("Case #"+cased+": "+b.pow(n).toString());
    13             cased++;
    14         }
    15     }
    16 }
    View Code
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  • 原文地址:https://www.cnblogs.com/demodemo/p/4678367.html
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