Sword offer - children's game (the last remaining number in the circle)

 

Topic description

 

Every year on Children's Day, Niu Ke will prepare some small gifts to visit the children in the orphanage, and this year is the same. As a senior veteran of Niu Ke, 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 calls 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, and continuing 0...m -1 count.... 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)

 

Problem solving ideas

 

This problem is a typical Joseph ring problem. Consider using the function f(n) to represent the last remaining number in the n numbers from 0 to n-1. The number to be removed is (m-1)%n, because the next number is Starting from the m%n number, the corresponding relationship between the number in f(n-1) and the number in f(n) is

f（n-1）         f（n）

0                m%n

1               (m+1)%n

…               …

n-2             (m-2)%n

The relationship between the two can be summarized: f(n) = (f(n-1) + m)% n When n=1, the remaining numbers in the circle must be 0, so the recurrence relationship is as follows:

f（n）= 0                     n=1

       （f（n-1）+ m）% n     n>1

 

code

 

 1 class Solution {
 2 public:
 3     int LastRemaining_Solution(int n, int m)
 4     {
 5         if(n<1||m<1)
 6             return -1;
 7         int l=0;
 8         for(int i=2;i<=n;i++)
 9             l=(l+m)%i;
10         return l;
11     }
12 };