Hdu 5241 Friends【规律+高精度乘法】

最新推荐文章于 2018-04-22 23:15:46 发布

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最新推荐文章于 2018-04-22 23:15:46 发布

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本文介绍了一道关于语言掌握情况的编程题，主人公精通N种语言，有九位朋友掌握的语言都是其子集，根据特定的关系，求解可能的语言掌握情况总数。通过观察规律，发现结果随N增加呈指数级增长。

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Friends

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1451 Accepted Submission(s): 691

Problem 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≤20) denoting the number of test cases.

For each test case, the first line contains an integer

0≤n≤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

题目大意：

主人公会N种语言，现在有九个小伙伴，已知他们之间的子集关系，问小伙伴们能够掌握的语言的情况数。

思路：

本地打表，找规律发现n每加1，结果增长2^5倍。

那么写一个高精度即可。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int n;
int a[20000];
void Yunsuan(int num)
{
    if(num>=0)
    {
        for(int i=0;i<5000;i++)a[i]*=32;
        for(int i=0;i<5000;i++)
        {
            if(a[i]>=10)
            {
                a[i+1]+=a[i]/10;
                a[i]%=10;
            }
        }
    }
    if(num==n-1)
    {
        int pos;
        for(int i=5000;i>=0;i--)
        {
            if(a[i]!=0)
            {
                pos=i;
                break;
            }
        }
        for(int i=pos;i>=0;i--)
        {
            printf("%d",a[i]);
        }
        printf("\n");
    }
}
int main()
{
    int kase=0;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        memset(a,0,sizeof(a));
        a[0]=1;
        printf("Case #%d: ",++kase);
        for(int i=-1;i<n;i++)
        {
            Yunsuan(i);
        }
    }
}