E. Tutorial Groupings (DP + Thinking)

Title:

gives you $N$ number, then let you group, each group has at most$s$ elements, and the entire sequence is required to be strictly increasing, that is,$T_i$The largest element of the group is strictly less than $T_{i+1}$the smallest element in . And the range in a group is at most $k$ . Ask the number of programs? Modulo 1e9+7.

analyze:

First we look at the data range n is $10000$ , s is$100$ ,
then we can use$dp\left[i\right]\left[j\right]$ represents the i-th position, the number of solutions ending in a group of length j.

Two cases:

• $j=1.$ Then we can think of$dp\left[i\right]\left[1\right]$ represents the i-th position, and he ends up as a group. What state can he transfer from there? He can get it from$dp[i-1]\left[1\right],dp[i-1]\left[2\right]......dp[i-1]\left[s\right]$ Transfer over, which is equivalent to starting a new stove.

• Then we look at $j!=1$，$dp\left[i\right]\left[j\right]$ is definitely from$dp[i-1][j-1]$ , and then add one more last time. If you can put it down, you can take over the previous state. If you can't, it means you can't reach it.
  Whether it can be put down depends on whether it is satisfied: whether the range within this group is less than or equal to k. That is, whether the difference between the largest and the smallest meets the conditions. If the condition is met:$dp\left[i\right]\left[j\right]=dp[i-1][j-1];$ if not$dp\left[i\right]\left[j\right]=0;$

///苟利国家生死以，岂因祸福避趋之。
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn = 2e5 + 7;
const int mod = 1e9 + 7;
const double pi = 3.14159265358979;
ll n,k,s;
ll a[maxn];
ll dp[20000][200];
ll sum;
int main()
{
    
    
    cin >> n >> k >> s;
    for(int i = 1; i <= n; i++)
    {
    
    
        cin >> a[i];
    }
    sort(a + 1, a + 1 + n);

    for(int i = 1; i <= n; i++) dp[i][1] = 1;    
	///直接dp[1][1]=1也行。
    for(ll i = 2; i <= n; i++)
    {
    
    
        sum=0;
        for(ll j=1;j<=s;j++) sum+=dp[i-1][j],sum%=mod;

        dp[i][1]=sum;

        for(ll j=2;j<=min(i,s);j++){
    
    ///j 是i位置长度 那么 i-1位的长度为j-1
            if(a[i]-a[i-j+1]<=k){
    
    
                dp[i][j]=dp[i-1][j-1];
            }else {
    
    
                dp[i][j]=0;
            }
        }
    }
    sum=0;
    for(int i=1;i<=s;i++){
    
    
        sum+=dp[n][i];
        sum%=mod;
    }
    cout<<sum<<endl;
    return 0;
}