杭电1592Half of and a Half（大数）

喜欢ヅ旅行 2022-08-17 15:16 95阅读 0赞

# 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;
    }