D - Half of and a Half 大数

D - Half of and a Half

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1592

Description

Gardon bought many many chocolates from the A Chocolate Market (ACM). When he was on the way to meet Angel, he met Speakless by accident.
"Ah, so many delicious chocolates! I'll get half of them and a half!" Speakless said.
Gardon went on his way, but soon he met YZG1984 by accident....
"Ah, so many delicious chocolates! I'll get half of them and a half!" YZG1984 said.
Gardon went on his way, but soon he met Doramon by accident....
"Ah, so many delicious chocolates! I'll get half of them and a half!" Doramon said.
Gardon went on his way, but soon he met JGShining by accident....
"Ah, so many delicious chocolates! I'll get half of them and a half!" JGShining said.
.
.
.
After had had met N people , Gardon finally met Angel. He gave her half of the rest and a half, then Gardon have none for himself. Could you tell how many chocolates did he bought from ACM?

 

Input

Input contains many test cases.
Each case have a integer N, represents the number of people Gardon met except Angel. N will never exceed 1000;

 

Output

For every N inputed, tell how many chocolates Gardon had at first.

 

Sample Input

2

 

Sample Output

7

 1 #include<cstdio>

 2 #include<string.h>

 3 using namespace std;

 4 int ans[1000][510];

 5 int main()

 6 {

 7     //memset(ans,0,sizeof(ans));

 8     int n,i,j;

 9     while(scanf("%d",&n)!=EOF)

10     {

11 

12         //i=300;

13 memset(ans,0,sizeof(ans));

14 ans[500]=1;

15         while(n--)

16         {

17             int c=0;

18             int s=0;

19 

20             for(i=500;i>=0;i--)

21             {

22                 if(i==500)

23                  s=ans[i]*2+1+c;

24                  else s=ans[i]*2+c;

25                 ans[i]=s%10;

26                 c=s/10;

27 

28             }

29         }

30         for(i=0;i<=500;i++) if(ans[i]!=0) break;

31         j=i;

32         for(j=i;j<=500;j++) printf("%d",ans[j]);

33         printf("\n");

34     }

35     return 0;

36 }