# 2393. Count Strictly Increasing Subarrays
###### tags: `Leetcode` `Medium` `Dynamic Programming`
Link: https://leetcode.com/problems/count-strictly-increasing-subarrays/description/
## 思路 $O(N)$ $O(N)$

所有的```dp[i]```加在一起就是答案
空间复杂度可以简化为$O(1)$
## Code

class Solution {
public long countSubarrays(int[] nums) {
long[] dp = new long[nums.length];
Arrays.fill(dp, 1);
long ans = dp[0];
for(int i=1; i<nums.length; i++){
if(nums[i]>nums[i-1]) dp[i] = dp[i-1]+1;
ans += dp[i];
}
return ans;
}
}