51nod 1073 Joseph Ring (mathematical recursion)

1073 Joseph Ring Reference time limit: 1 seconds space limit: 131072 KB score: 0 Difficulty: Basic collection Attention N the person sits into a circle (numbered 1-n), from the 1th person starts to count, the number to K of the person out, the back of the person from 1 began counting off. Ask the number of the last remaining person. For example: N = 3,k = 2. Number 2nd, then number 1th, and the last one is number 3rd. Input

2 numbers n and K, which represent n individuals, count to K to dequeue. (2 <= N, K <= 10^6)

Output

The number of the last remaining person

Input example

3 2

Output example

3

I feel that maths problem is very difficult ...

The analysis process is this, assuming that the first round of elimination is numbered k-1 (numbering is starting from 0), then there will be a first round and the second round of mapping relationship.

K----------------->0

K+1-------------->1

...

N-1--------------->n-k-1

0------------------>n-k

...

K-2--------------->n-2 (the first round on the left and the second round on the right)

Set X as the number in the second round, according to the above mapping, there will be such a relationship (X+k)%n. The reason is that N is the remainder because it is from the second round n-1 personal mapping to the first round of n individuals.

And k=m%n, bring into the formula can get (x+m)%n.

And so pushed down, from the second round to the third round, it is (x+m)% (n-1), so recursion until the end.

Can think of the last round of time, there is only one person, then he is the winner, numbered 0.

So according to the above relationship, back, get the last round numbered 0 people in the first round of the number, is the answer.

#include <iostream>
#include <cstdio>
#include <cstring>
#define  max_n 1000005
using namespace std;
int main ()
{
    int n,k;
    int f[max_n];
    while (cin>>n>>k)
    {
        f[1]=0;
        for (int i=2;i<=n;i++)
        {
            f[i]= (f[i-1]+k)%i;
        }
        printf ("%d\n", f[n]+1);
    }
    return 0;
}