DP专题训练5

数组切分

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define endl "\n"
/*
1 3 2 4
1:
{1}dp[1]+=dp[0]
2:
{1}{3}dp[2]+=dp[0]
3:
{1}{3}{2}dp[3]+=dp[2]
{1}{3,2}dp[3]+=dp[1]
{1,3,2}dp[3]+=dp[0]
4:
{1}{3}{2}{4}|{1}{3,2,4}|{1,3,2,4}|dp[4]+=dp[3]
{1}{3,2,4}dp[4]+=dp[1]
{1,2,3,4}dp[4]+=dp[0];
dp[i]表示长度为i的划分数 当[j,i]这段区间内满足连续自然数段(即mx-mi+1=i-j+1),则可以由dp[j-1]转移来
*/
const int N = 1e4 + 10;
const int mod = 1e9 + 7;
int n;
int a[N],dp[N];
void solve(){
	cin >> n;
	dp[0] = 1;
	for(int i = 1; i <= n ; i ++) cin >> a[i];
	for(int i = 1; i <= n ; i ++){
		int mx = a[i], mn = a[i];
		for(int j = i; j >= 1 ; j --){
			mx = max(mx,a[j]), mn = min(mn,a[j]);
			if(mx - mn + 1 == i - j + 1)
				dp[i] = (dp[i] + dp[j-1]) % mod;
		}
	}
	cout << dp[n];
}
signed main(){
	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	int _ = 1;
	//cin >> _;
	while(_--) solve();
	return 0;
}