Sum of factor numbers of interval product

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The question ：
Given a length of $n$ Array of $a$, Each number in the array is not greater than $3$ The positive integer . It is known that the weight of an interval is the number of factors of the product of all numbers in the interval , Find the sum of the weights of all intervals . The answer is right $1e9+7$ modulus .

Data range ：
$1\le n\le 10^5$
$1\le a_i\le 3$

Answer key ：
a number $x$ The number of factors for is ：
take $x$ Split into the product form of all prime factors , Count the number of each germplasm factor , The first $i$ A quality factor $p_i$ The quantity of is $cn{t}_{p_i}$
So count $x$ The number of factors for is ：$\prod _{i=1}(cn{t}_{p_i}+1)$

$pre{2}_i=\sum _{j=1}^i[a_j=2]$ ,$pre{3}_i=\sum _{j=1}^i[a_j=3]$

$ppre{2}_i=\sum _{j=1}^{i}pre{2}_j$,$ppre{3}_i=\sum _{j=1}^{i}pre{3}_j$

$mul23_i=\sum _{j=1}^{i}pre{2}_j\times pre{3}_j$

For the interval weight in the topic , Equivalent to $(pre2+1)\times (pre3+1)$.

So the question is ：$\sum _{r=1}^n\sum _{l=1}^r(pre{2}_r-pre{2}_{l-1}+1)\times (pre{3}_r-pre{3}_{l-1}+1)$

Simplification ：
$$\begin{gathered} (pre{2}_r-pre{2}_{l-1}+1)\times (pre{3}_r-pre{3}_{l-1}+1) \\ =pre{2}_r\times pre{3}_r-pre{2}_r\times pre{3}_{l-1}+pre{2}_r-pre{2}_{l-1}\times pre{3}_r+pre{2}_{l-1}\times pre{3}_{l-1}-pre{2}_{l-1}+pre{3}_r-pre{3}_{l-1}+1 \\ =(pre{2}_r\times pre{3}_r+pre{2}_r+pre{3}_r+1{)}_{①}+(pre{2}_{l-1}\times pre{3}_{l-1}{)}_{②}-(pre{2}_{l-1}\times (pre{3}_r+1)+pre{3}_{l-1}\times (pre{2}_r+1){)}_{③} \end{gathered}$$

Sum the above three expressions respectively ：
① $$\begin{gathered} \sum _{r=1}^n\sum _{l=1}^r(pre{2}_r\times pre{3}_r+pre{2}_r+pre{3}_r+1) \\ =\sum _{r=1}^{n}r\times (pre{2}_r\times pre{3}_r+pre{2}_r+pre{3}_r+1) \end{gathered}$$

② $$\begin{gathered} \sum _{r=1}^n\sum _{l=1}^r(pre{2}_{l-1}\times pre{3}_{l-1}) \\ =\sum _{r=1}^{n}mul23_{r-1} \end{gathered}$$

③ $$\begin{gathered} \sum _{r=1}^n\sum _{l=1}^r-(pre{2}_{l-1}\times (pre{3}_r+1)+pre{3}_{l-1}\times (pre{2}_r+1)) \\ =-\sum _{r=1}^n\sum _{l=1}^r(pre{2}_{l-1}\times (pre{3}_r+1)+pre{3}_{l-1}\times (pre{2}_r+1)) \\ =-\sum _{r=1}^n(ppre{2}_{r-1}\times (pre{3}_r+1)+ppre{3}_{r-1}\times (pre{2}_r+1)) \end{gathered}$$

Can be in $O(n)$ Complete in time complexity .

Code ：

#include <bits/stdc++.h>
using namespace std;

const int mod = 1e9 + 7;
const int N = 100010;
int a[N], n;
int pre2[N], ppre2[N];
int pre3[N], ppre3[N];
int mul23[N];

void add(int& a, int b) {
    
	a = (a + b) % mod;
	if (a < 0) a += mod;
}

int main()
{
    
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
	for (int i = 1; i <= n; ++i) {
    
		pre2[i] = pre2[i - 1];
		pre3[i] = pre3[i - 1];
		
		if (a[i] == 2) {
    
			add(pre2[i], 1);
		} else if (a[i] == 3) {
    
			add(pre3[i], 1);
		}
		
		ppre2[i] = ppre2[i - 1];
		ppre3[i] = ppre3[i - 1];
		add(ppre2[i], pre2[i]);
		add(ppre3[i], pre3[i]);
		
		mul23[i] = mul23[i - 1];
		add(mul23[i], 1ll * pre2[i] * pre3[i] % mod);
	}
	
	int ans = 0;
	for (int i = 1; i <= n; ++i) {
    
		int temp = 0;
		add(temp, 1ll * pre2[i] * pre3[i] % mod);
		add(temp, pre2[i]);
		add(temp, pre3[i]);
		add(temp, 1);
		add(ans, 1ll * i * temp % mod);

		add(ans, -1ll * (1 + pre3[i]) * ppre2[i - 1] % mod);
		add(ans, -1ll * (1 + pre2[i]) * ppre3[i - 1] % mod);
		add(ans, mul23[i - 1]);
	}
	
	printf("%d\n", ans);
	
	return 0;
}