Coconuts, Revisited

　　　　　　　　　　Coconuts, Revisited

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


题意：总共有n个椰子，（自己以为是可可豆）有m个人，到了晚上，每个人都会去把n-1（给猴子一个）除以m，再拿走自己的一份，最后剩的还要是m的倍数，求最大m
如：25……先减一……=24……再除以m（=3）……=8……第一个人拿走了8个，剩下了16个……第二个人也要拿……-1再除3为5……即第二个人拿走了5个……还剩10个……-1再除3为3……即第三个人拿走了3个……还剩6个……
没人拿了，而且剩下了3的倍数
tip：模拟，下一次剩的椰子等于这一次减去这个人拿的再减一；即x=x-(x-1)/i-1

#include<iostream>

using namespace std;

int solve(int n)
{
    int i,j,x;
    for(i=10;i>1;i--)
    {
        x=n;
        j=i;
        while(j)
        {
            if((x-1)%i!=0)
                break;
            x=x-(x-1)/i-1;
            j--;
        }
        if(j==0&&x%i==0)
            return i;
    }
    return 0;
}

int main()
{
    int c,p;
    while(cin>>c)
    {
        if(c==-1)
            break;
        cout<<c<<" coconuts, ";
        p=solve(c);
        if(p)
            cout<<p<<" people and 1 monkey"<<endl;
        else
            cout<<"no solution"<<endl;
    }
    return 0;
}