hdu 5138 CET-6 test（BestCoder Round #21）-优快云博客

本文链接：https://blog.youkuaiyun.com/caduca/article/details/41786519

本文介绍了一种基于艾宾浩斯遗忘曲线的记忆方法帮助考生高效复习CET-6考试词汇，通过合理安排复习时间，提高学习效率。

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CET-6 test

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 90 Accepted Submission(s): 74

Problem Description

Small W will take part in CET-6 test which will be held on December

20th. In order to pass it he must remember a lot of words.
He remembers the words according to Ebbinghaus memory curve method.
He separates the words into many lists. Every day he picks up a new list, and in the next

1st,2nd,4th,7th,15th day, he reviews this list.
So every day he has many lists to review. However he is so busy, he does not know which list should be reviewed in a certain day. Now he invites you to write a program to tell him which list should to be reviewed in a certain day.
Lists are numbered from 1. For example list 1 should be reviewed in the

2nd,3rd,5th,8th,16th day.

Input

Multi test cases (about 100), every case contains an integer n in a single line.
Please process to the end of file.

[Technical Specification]

2≤n≤100000

Output

For each n，output the list which should be reviewed in the

nth day in ascending order.

Sample Input

2
100

Sample Output

1
85 93 96 98 99

求第n天要复习的内容，要复习前1,2，4,7,15天的内容，其实只要用个数组存下要减的天数并判断是否大于0就可以了，当时脑抽直接分类讨论了，这样容易错。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        if(n<3)
        {
            printf("%d\n",n-1);
        }
        else if(n<5)
        {
            printf("%d %d\n",n-2,n-1);
        }
        else if(n<8)
        {
            printf("%d %d %d\n",n-4,n-2,n-1);
        }
        else if(n<16)
        {
            printf("%d %d %d %d\n",n-7,n-4,n-2,n-1);
        }
        else
        {
            printf("%d %d %d %d %d\n",n-15,n-7,n-4,n-2,n-1);
        }
    }
    return 0;
}