HDU1592(递推+大数处理)

Half of and a Half

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 792    Accepted Submission(s): 350

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
  
  

 

递推+大数处理

#include<iostream>
using namespace std;
int a[100];

int main() {  int n,i,j,temp;  int len;  while(scanf("%d",&n)!=EOF)  {   memset(a,sizeof(a));   len=0;   a[len]=1;   for(i=1;i<=n;i++)   {    temp=0;    for(j=0;j<=len;j++)    {     if(j==0)      a[j]=(a[j]+0.5)*2;     else a[j]=a[j]*2+temp;     temp=a[j]/10000;     a[j]=a[j]%10000;    }    if(temp)    {     len++;     a[len]=temp;    }   }   printf("%d",a[len]);   for(i=len-1;i>=0;i--)    printf("%04d",a[i]);   printf("\n");     }  return 0; }