"Children's Game (Last Remaining Numbers in the Circle)" from "The Sword Offer"

Topic description

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Every year on Children's Day, teachers will prepare some small gifts to visit the children in the orphanage, and this year is the same. As a senior veteran of the teacher, HF naturally prepared some small games. Among them, there is a game like this: First, let the children form a big circle. Then, he randomly assigned a number m, and let the children numbered 0 start counting. Every time the child who shouts m-1 will go out to sing a song, and then he can choose any gift from the gift box, and will not return to the circle, starting from his next child, continue 0...m-1 Report the number.... and so on.... until the last child left, you can not perform, and get the valuable "Detective Conan" collector's edition (limited places!! ^_^). Please try to think, which child will get this gift? (Note: Children are numbered from 0 to n-1)

Code

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//我的思路 是用队列。
//每次找到第(m-1)%n 的扔掉。
//最后剩下的就是答案了。
class Solution {
public:
    int LastRemaining_Solution(int n, int m)
    {    queue<int> Q;
     if (n <= 0 || m <= 0 ){
         return -1 ;
     }
     for(int i = 0;i< n ;i++){
         Q.push(i);   
        }
        while(Q.size()!= 1){
            int cent = (m - 1) % n;
            for(int j = 0;j<cent ;j++){
                Q.push(Q.front());
                Q.pop();
            }
            Q.pop();
            --n;
        }
    return Q.front();
    }
};