Half of and a Half

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1179    Accepted Submission(s): 539

Problem 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<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 struct node
 6 {
 7     int a[1100];
 8 }s[1010];
 9 int main()
10 {
11     int i,j,t,n;
12     s[0].a[0]=1;
13     for(i=1;i<=1000;i++)
14     {
15         int r=0;
16         for(j=0;j<1000;j++)
17         {
18             s[i].a[j]=s[i-1].a[j]*2+r;
19
20             r=s[i].a[j]/10;
21
22             s[i].a[j]%=10;
23
24         }
25         t=0;
26         while(s[i].a[t]+1>9)
27         {
28             s[i].a[t]=0;
29             t++;
30         }
31         s[i].a[t]+=1;
32     }
33     while(scanf("%d",&n)!=EOF)
34     {
35         for(i=999;i>=0,s[n].a[i]==0;i--);
36         for(j=i;j>=0;j--)
37             printf("%d",s[n].a[j]);
38         printf("\n");
39
40     }
41     return 0;
42 }