杭电5138-CET-6 test

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
思路：其实这道题想明白了，很简单，就是一个记忆曲线，用输入的数减去15,7,4,1即可

#include<iostream>#include<stdio.h>using namespace std;int main(){int n;while(cin>>n){if(n>15){printf("%d %d %d %d %d\n",n-15,n-7,n-4,n-2,n-1);}else if(n>7&&n<=15)printf("%d %d %d %d\n",n-7,n-4,n-2,n-1);else if(n==2)printf("%d\n",n-1);else if(n>2&&n<=4)printf("%d %d\n",n-2,n-1);else if(n>4&&n<=7)printf("%d %d %d\n",n-4,n-2,n-1);}return 0;}

还有一种简单的方法

#include<iostream>#include<stdio.h>using namespace std;int main(){    int n;    while(cin>>n)    {        if(n>15)            printf("%d ",n-15);        if(n>7)            printf("%d ",n-7);        if(n>4)            printf("%d ",n-4);        if(n>2)            printf("%d ",n-2);        if(n>1)            printf("%d\n",n-1);    }    return 0;}