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

[b]D. Duff in Beach[/b]

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

10

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
[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 = 1;
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=1;i<=n;i++)
{
scanf("%d",&a[i]);
b[i]=a[i];
}
sort(a+1,a+1+n);
for(int i=1;i<=n;i++)
HH[a[i]] = i;
for(int i=1;i<=n;i++)
dp[get(i,1)]=1;
for(int j=2;j<=k;j++)
{
int tot = 1;
for(int i=1;i<=n;i++)
{
int kk = upper_bound(a+1,a+n+1,a[i])-(a+1);
while(tot<=kk)
{
sum[j-1] += dp[get(tot,j-1)];
sum[j-1] %= mod;
tot++;
}
dp[get(i,j)] = sum[j-1];
}
}
for(int i=1;i<=n;i++)
{
sum[k] += dp[get(i,k)];
sum[k] %= mod;
}
long long ans = 0;
for(int i=1;i<=k&&i<=l/n;i++)
{
ans += (long long)((l/n-i+1)%mod)*sum[i];
ans %= mod;
}
if(l%n==0)
{
cout<<ans<<endl;
return 0;
}
for(int i=1;i<=l%n;i++)
{
for(int j=1;j<=k&&j<=(l/n+1);j++)
{
ans+=dp[get(HH[b[i]],j)];
ans%=mod;
}
}
cout<<ans<<endl;
}