Codeforces Round #326 (Div. 2) D. Duff in Beach dp

D. Duff in Beach

Time Limit: 1 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/588/problem/D

Description

While Duff was resting in the beach, she accidentally found a strange array b0, b1, ..., bl - 1 consisting of l positive integers. This array was strange because it was extremely long, but there was another (maybe shorter) array, a0, ..., an - 1 that b can be build from a with formula: bi = ai mod n where a mod b denoted the remainder of dividing a by b.

Duff is so curious, she wants to know the number of subsequences of b like bi1, bi2, ..., bix (0 ≤ i1 < i2 < ... < ix < l), such that:

1 ≤ x ≤ k
For each 1 ≤ j ≤ x - 1,
For each 1 ≤ j ≤ x - 1, bij ≤ bij + 1. i.e this subsequence is non-decreasing.

Since this number can be very large, she want to know it modulo 109 + 7.

Duff is not a programmer, and Malek is unavailable at the moment. So she asked for your help. Please tell her this number.

Input

The first line of input contains three integers, n, l and k (1 ≤ n, k, n × k ≤ 106 and 1 ≤ l ≤ 1018).

The second line contains n space separated integers, a0, a1, ..., an - 1 (1 ≤ ai ≤ 109 for each 0 ≤ i ≤ n - 1).

Output

Print the answer modulo 1 000 000 007 in one line.

Sample Input

3 5 3
5 9 1

Sample Output

HINT

题意

给你n,l,k，然后就是告诉你b有l长度，其中是由a不停重复得到的

然后问你一共有多少个满足条件的序列存在

条件如下：

1.这个序列的长度大于等于1，小于等于k

2.这个序列在每一个块中只能选择一个数，并且都必须选择在连续的块中

3.这个序列是单调不降的

题解：

dp，由于n*k<=1e6，所以我们简单的推断是dp[n][k]的，表示第i个数长度为k的序列有多少个

其实这道题和分块差不多，在块外就用dp瞎转移

块内就暴力就好了……

只要把这个数据过了就好了

3 100 3

1 1 1

注意是upper_bound这个东西

代码：

#include<stdio.h>

#include<iostream>

#include<algorithm>

#include<map>

using namespace std;

#define maxn 1000005

#define mod 1000000007

int a[maxn];

int b[maxn];

long long l;

int k,n;

long long dp[maxn];

long long sum[maxn];

map<int,int> HH;

map<pair<int,int>,int> H;

int tot = ;

int get(int x,int y)

{

    if(H[make_pair(x,y)])

        return H[make_pair(x,y)];

    H[make_pair(x,y)]=tot++;

    return H[make_pair(x,y)];

}

int main()

{

    scanf("%d%lld%d",&n,&l,&k);

    for(int i=;i<=n;i++)

    {

        scanf("%d",&a[i]);

        b[i]=a[i];

    }

    sort(a+,a++n);

    for(int i=;i<=n;i++)

        HH[a[i]] = i;

    for(int i=;i<=n;i++)

        dp[get(i,)]=;

    for(int j=;j<=k;j++)

    {

        int tot = ;

        for(int i=;i<=n;i++)

        {

            int kk = upper_bound(a+,a+n+,a[i])-(a+);

            while(tot<=kk)

            {

                sum[j-] += dp[get(tot,j-)];

                sum[j-] %= mod;

                tot++;

            }

            dp[get(i,j)] = sum[j-];

        }

    }

    for(int i=;i<=n;i++)

    {

        sum[k] += dp[get(i,k)];

        sum[k] %= mod;

    }

    long long ans = ;

    for(int i=;i<=k&&i<=l/n;i++)

    {

        ans += (long long)((l/n-i+)%mod)*sum[i];

        ans %= mod;

    }

    if(l%n==)

    {

        cout<<ans<<endl;

        return ;

    }

    for(int i=;i<=l%n;i++)

    {

        for(int j=;j<=k&&j<=(l/n+);j++)

        {

            ans+=dp[get(HH[b[i]],j)];

            ans%=mod;

        }

    }

    cout<<ans<<endl;

}

巴特西