F - Infinite Sequence

Time limit : 2sec / Memory limit : 256MB

Score : 1000 points

Problem Statement

How many infinite sequences a1,a2,… consisting of {1,…,n} satisfy the following conditions?

• The n-th and subsequent elements are all equal. That is, if n≤i,j, ai=aj.
• For every integer i, the ai elements immediately following the i
  -th element are all equal. That is, if i<j<k≤i+ai, aj=ak.

Find the count modulo 109+7.

Constraints

• 1≤n≤106

Input

Input is given from Standard Input in the following format:

n

Output

Print how many sequences satisfy the conditions, modulo 109+7.

Sample Input 1

2

Sample Output 1

4

The four sequences that satisfy the conditions are:

• 1,1,1,…
• 1,2,2,…
• 2,1,1,…
• 2,2,2,…

Sample Input 2

654321

Sample Output 2

968545283

各個元素範圍在[1,N]，第i個元素後面a[i]個元素都要一樣，第N個元素及其往後的數字都一樣。

思路：發現1是比較特殊的一個數，定義dp[i]為空出前n-i個位置時的解，當第一個數字為1時，顯然dp[i] = dp[i-1]，當第一個數字不為1且第二個數字不為1時，顯然數列為abbbbbb...形式，有(n-1)*(n-1)種，當第一個數字不為1且第二個數字為1時，有dp[i] += dp[i-3] + dp[i-4] +...+dp[1] + (n-i+2)種。

# include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const LL mod = 1e9+7;
LL dp[1000003];
int main()
{
    LL n, sum=0;
    scanf("%lld",&n);
    dp[1] = n;
    dp[2] = (n*n)%mod;
    for(int i=3; i<=n; ++i)
    {
        sum = (sum + dp[i-3])%mod;
        dp[i] = (dp[i-1] + sum + (n-1)*(n-1)%mod + n-i+2)%mod;
    }
    printf("%lld\n",dp[n]);
    return 0;
}