Coconuts, Revisited（递推+枚举+模拟）

Description

The short story titled Coconuts, by Ben Ames Williams, appeared in the Saturday Evening Post on October 9, 1926. The story tells about five men and a monkey who were shipwrecked on an island. They spent the first night gathering coconuts. During the night, one man woke up and decided to take his share of the coconuts. He divided them into five piles. One coconut was left over so he gave it to the monkey, then hid his share and went back to sleep.
Soon a second man woke up and did the same
thing. After dividing the coconuts into five piles, one coconut was left
over which he gave to the monkey. He then hid his share and went back
to bed. The third, fourth, and fifth man followed exactly the same
procedure. The next morning, after they all woke up, they divided the
remaining coconuts into five equal shares. This time no coconuts were
left over.
An obvious question is ``how many coconuts
did they originally gather?" There are an infinite number of answers,
but the lowest of these is 3,121. But that's not our problem here.
Suppose we turn the problem around. If we know the number of coconuts
that were gathered, what is the maximum number of persons (and one
monkey) that could have been shipwrecked if the same procedure could
occur?

Input

The input will consist of a sequence of integers, each representing the
number of coconuts gathered by a group of persons (and a monkey) that
were shipwrecked. The sequence will be followed by a negative number.

Output

For each number of coconuts, determine the largest number of persons
who could have participated in the procedure described above. Display
the results similar to the manner shown below, in the Sample Output.
There may be no solution for some of the input cases; if so, state that
observation.

Sample Input

25
30
3121
-1

Sample Output

25 coconuts, 3 people and 1 monkey
30 coconuts, no solution
3121 coconuts, 5 people and 1 monkey

题目大意：还是人和猴子分桃子，不过猴子只有一个。和UVA-10726Coco Monkey不同的是，这道题已知的是桃子数，让求可能的最多人数。
题目解析：将递推的过程反过来，枚举人数，模拟分桃子的过程。人数不会太多。

代码如下：

 # include<iostream>
 # include<cstdio>
 # include<set>
 # include<vector>
 # include<fstream>
 # include<cstring>
 # include<algorithm>
 using namespace std;
 const int N=;
 bool ok(int ss,int n)
 {
     int t=ss;
     while(ss--){
         if((n-)%t)
             break;
         n=(n-)/t*(t-);
     }
     if(ss==-&&(n%t==))
         return true;
     return false;
 }
 int main()
 {
     int n,i;
     while(scanf("%d",&n))
     {
         if(n==-)
             break;
         int ans=;
         for(i=;i<;++i){
             if(i*(i-)>=n)
                 break;
             if(ok(i,n)){
                 ans=i;
             }
         }
         printf("%d coconuts, ",n);
         if(ans>){
             printf("%d people and 1 monkey\n",ans);
         }else
             printf("no solution\n");
     }
     return ;
 }