约瑟夫环(Josephuse)

• 时间复杂度是O(n);
• 空间复杂度O(1)

int LastRemaining(unsigned int n,unsigned int m){    if(n<1||m<1){        return -1;    }    int last=0;    for (int i=2; i<=n; i++) {        last=(last+m)%i;    }    return last;}