杭电1592 Half of and a Half 大数

Half of and a Half

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1022    Accepted Submission(s): 452

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
  

 

Author

DYGG

 

Source

HDU “Valentines Day” Open Programming Contest 2007-02-14

 

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公式很好推，2^（n+1）-1,附大数代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[1100][1100],i,j,k,l,m,n;
void ac()
{
	memset(a,0,sizeof(a));
	a[0][0]=1;
	for(i=1;i<1001+1;i++)
	{
		int flag=0;//进wei 
		for(j=0;j<1010;j++)
		{
			a[i][j]=a[i-1][j]*2+flag;
			flag=a[i][j]/10;
			a[i][j]%=10;
		}
	}
}
int main()
{
	ac();
	while(scanf("%d",&n)!=EOF)
	{
		for(i=1010-1;i>=0;i--)
		if(a[n+1][i])
		break;
		for(;i>0;i--)
		printf("%d",a[n+1][i]);
		printf("%d\n",a[n+1][i]-1);//那么问题来了，为什么没有考虑最后一位为0 的情况 
	}//那样的话，就成大数减了，因为2的次方最后一位只会是2 4 8 6，还有一个1. 
}