Array-11. Monkey selection King (20)

Array-11. Monkey selection King (20) time limit MS
Memory Limit 65536 KB
Code length limit 8000 B
Procedures for the award of questions StandardAuthor Xu Jianchun (Zhejiang University)

A group of monkeys will choose the new Monkey King. The new Monkey King's choice is: Let n only the candidate monkeys in a circle, from a position from a sequence numbered 1-n. Starting from the 1th number, each round from 1 to report 3, where the report 3 of the monkeys are out of the circle, and then from the next close to a monkey start the same count. So constantly circulating, the last one left is chosen as the Monkey King. What is the original number of monkeys elected Monkey King?

Input format:

The input gives a positive integer n (<=1000) in a row.

Output format:

Prints the number of the selected monkey in a row.

Input Sample:

11

Sample output:

7

The following ideas are wrong, beginning with the erase, but found that the iterator will be pumped//later with the turn to 3 of the element into 0, found also not, because it will involve the repetition into a 3 case//#include <iostream>//#include <         list>//using namespace std;//int sum (list<int> num)//{/int s = 0;//for (auto I:num)//{//     if (i = = 0)//++s;//}//return s;//}//int main ()//{//int n;//cin >> n;// List<int> num;//for (int i = 1; I <= N; ++i)//{//Num.push_back (i);//}//int flag = 0;             Auto it = Num.begin ();//while (SUM (num)! = N-1)//{//if (it = = Num.end ())///{//             it = Num.begin ();//}//++flag;//if (flag = = 3)//{//*it = 0;//   Flag = 0;//}//++it;         Next iter////cout << *it << endl;//}//for (auto I:num)//{//if (I! = 0)// {//cout << i << endl;//}//}//return 0;//}

Finally change the erase, you can use the erase!

#include <iostream> #include <list>using namespace Std;int main () {    int N;    Cin >> N;    list<int> num;    for (int i = 1; I <= N; ++i)    {        num.push_back (i);    }    int flag = 0;        Symbol of the    auto it = Num.begin ();    while (Num.size () > 1)    {        ++flag;        if (flag = = 3)        {            it = num.erase (it);//Note It is invalid after deletion and cannot go to ++!!!            //erase will return the next pointer to the deleted element            flag = 0;        }        if (flag! = 0)      //At first there is a mistake, due to the deletion of the save operation, resulting in the deletion of the words will automatically move to the next            ++it;   Next ITER        if (it = = Num.end ())        {            it = Num.begin ();        }    }    cout << *num.begin () << Endl;    return 0;}

Array-11. Monkey selection King (20)