Take turns taking the stone game

• Problem Description

Two smart enough people play the game of taking stones in turns. Whoever takes the last stone wins. They can only take 1, 3, 7 or 8 stones at a time. When writing a program to judge n stones, take first. Of people are winning or losing.

Input format:
an integer n whose value does not exceed 10000000.
Output format:
If the person who takes first wins, please output 1 in a single line, otherwise output 0.

Input sample:
Here are 3 groups of input.

1
10
300

Output sample:
The output corresponding to the above three sets of data are as follows:

1
1
0

• Problem interpretation

First of all, these two people are smart enough. If the first person takes first, he will try his best to win. But the second person is not a vegetarian, he will spare no effort to prevent the first person from winning. Since the number of stones is uncertain, we might as well try the water with a smaller number of stones first to see if we can find a pattern.

• Enumeration analysis

We assume that $P_1$、$P_2$It represents the first person and the second person respectively, and the first person takes first. Now let's enumerate several situations and see under what circumstances the first person will win.

A stone:

Note: The first person wins.

Two stones:

Note: The first person loses.

Three stones:

The first situation:

Note: The first person wins.

The second case:

Note: The first person wins.

Four stones:

The first situation:

Note: The first person wins.

The second case:

Note: The first person wins.

Five stones:

The first situation:

Note: The first person wins.

The second case:

Note: The first person wins.

The third case:

Note: The first person wins.

Continue to enumerate and find that when the number of stones is: 2 4 6 15 17 19 30 32 34 36 45 47 49 51..., the second person wins. Observing these data, we can find that when n%15==0 or 2 or 4 or 6, the second person wins, and the first person wins in other situations. In this way, we have found the law.

• Code

#include<iostream>
using namespace std;
int main()
{
    
    
	int n;
	while(cin>>n)
	{
    
    
	    if(n%15==0)
            cout<<0;
        else if(n%15==2)
            cout<<0;
        else if(n%15==4)
            cout<<0;
        else if(n%15==6)
            cout<<0;
        else
            cout<<1;
	}
	return 0;
}

• operation result