題目描述

Slot machines are popular game machines in casinos. The slot machine we are considering has six places where a figure appears. By combination of figures, one may earn or lose money. There are ten kinds of figures, so we will represent a figure with a number between 0 and 9. Then we can use a six-digit number w = w1w2w3w4w5w6 where 0 ≤ w1, w2, w3, w4, w5, w6 ≤ 9 to represent one possible outcome of the slot machine.

Figure I.1. The layout of a slot machine.

Old slot machines were made up with mechanical components, but nowadays they were replaced by PC-based systems. This change made one critical flaw: they are based on pseudo-random number generators and the outcome sequences of a slot machine are periodic. Let T[i] be the i-th outcome of a slot machine. At first, there is a truly random sequence of length k,T[1],T[2],…,T[k]. Then there exists one positive number such that T[i+p] = T[i]for all possible values of i(>k). Once an attacker can find out the exact values of k and p, he or she can exploit this fact to beat the casino by betting a lot of money when he or she knows the outcome with a good combination in advance. For example, you have first six numbers of outcome sequences: 612534, 3157, 423, 3157, 423, and 3157. Note that we can remove first 0’s. Therefore, 3157 represents 003157 and 423 represents 000423. You want to know its tenth number. If you know the exact values of k and p, then you can predict the tenth number. However, there are many candidates for k and p: one extreme case is k=5 and p=1, and another is k=0 and p=6. The most probable candidate is the one where both k and p are small. So, our choice is the one with the smallest k+p. If there are two or more such pairs, we pick the one where p is the smallest. With our example, after some tedious computation, we get k=1 and p=2. Assume that you have n consecutive outcomes of a slot machine, T[1], T[2], …, T[n]. Write a program to compute the values of k and p satisfying the above-mentioned condition.

輸入

Your program is to read from standard input. The first line contains a positive integer n (1 ≤ n ≤ 1,000,000), representing the length of numbers we have observed up to now in the outcome sequence. The following line contains n numbers. Each of these numbers is between zero and 999,999.

輸出

Your program is to write to standard output. Print two integers k and p in one line.

樣例輸入

6
612534 3157 423 3157 423 3157

樣例輸出

1 2

題意:

連續刪除k 個數, 使得後面的數有迴圈節 ,且迴圈節長度為p

使得 k+p  儘可能的小,且當 k+p 一樣時,選擇p小的.

[思路]

  題意 轉化成  給你 一串數,   選擇後面的一些,求迴圈節..

求迴圈節.   KMP 演算法裡  next陣列  就可以  求  :   

迴圈節為 :  len - next[len];     

迴圈 次數 為 : if len %( len-next[len]) == 0  then  cot = len / ( len - next[len])

這裡用不到 迴圈次數, 只需要 迴圈節長度,  

只需要把陣列逆置一下,然後 列舉  k    求解.

[程式碼]

#include <bits/stdc++.h>
#include <stdio.h>
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define per(i,a,n) for(int i=n;i>=a;i--)

typedef long long ll;
const int maxn = 1e6+10;
const int mod =1e9+7;
const int inf = 0x3f3f3f3f;
using namespace std;

int a[maxn],vis[maxn];
int n;
int net[maxn];
void getfail(int p[],int n) //字串p自我匹配   
{
    int len = n;
    net[0]=net[1]=0;
    for(int i=1;i<len;i++)
    {
        int j=net[i];
        while(j&&p[i]!=p[j])
            j=net[j];
        if(p[i]==p[j])
            net[i+1]=j+1;//多匹配到了一個字元  
        else
            net[i+1]=0;//該字元配不上   
    }
}

int main(int argc, char const *argv[])
{
	int n;
	scanf("%d",&n);
	rep(i,0,n-1)
	{
		scanf("%d",&a[i]);
	}
	reverse(a,a+n); // 逆置
	getfail(a,n); // 求 next
	int mx = inf;
	int p = 0;
	int k = 0;
	rep(i,1,n) // 列舉 k
	{ 
		int lenp = (n-i+1 - net[n-i+1]); // 迴圈節..

		if( i+lenp < mx || (i+lenp)== mx && lenp < p )  // 
		{
			mx = i+lenp;
			p = lenp;
			k = i-1;
		}
	}
	printf("%d %d\n",k,p );

	return 0;
}