杭电1592Half of and a Half（大数）

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#杭电 #1592 #Half of and a Half #大数

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本文介绍了一道关于大数处理的编程题目，通过逆向思维和数据结构的应用，解决了不断分配一半加半个巧克力直至归零的问题。代码实现采用C++，展示了如何利用数组模拟大整数运算。

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Half of and a Half

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1023 Accepted Submission(s): 453

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

Sample Output

大数题，题意是见一个人分总数的一半+0.5，最后正好没有，问总共有多少！

要想最后一个人正好分完，倒数第二个人分完后应剩下1，这样一半+0.5才正好为0；

所以大数u！

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node
{
    int a[1100];
}s[1010];
int main()
{
    int i,j,t,n;
    s[0].a[0]=1;
    for(i=1;i<=1000;i++)
    {
        int r=0;
        for(j=0;j<1000;j++)
        {
            s[i].a[j]=s[i-1].a[j]*2+r;

            r=s[i].a[j]/10;

            s[i].a[j]%=10;

        }
        t=0;
        while(s[i].a[t]+1>9)
        {
            s[i].a[t]=0;
            t++;
        }
        s[i].a[t]+=1;
    }
    while(scanf("%d",&n)!=EOF)
    {
        for(i=999;i>=0,s[n].a[i]==0;i--);
        for(j=i;j>=0;j--)
            printf("%d",s[n].a[j]);
        printf("\n");

    }
    return 0;
}