Subsequence Problem Report Divisible by 3

Question link to the main
idea of ​​the question.
Give you a number string of length 50, and ask you how many subsequences constitute a number that can be divisible by 3. The
answer is $1e^9+7$ Modulo
input description
Input a string consisting of numbers, the length is less than or equal to 50
Output description
Output an integer
input example

132

Sample output

3

Dynamic programming linear dp
1. Determine the state $f\left[i\right]\left[j\right]$ :$The$ remainder at the end of$s$$[$$i$$]$ is$j$ , the attribute represents the number of schemes.
2. State transition$f[i+1][j]=f[i+1]\left[j\right]+f[i][j]$

Take a millet, take the sample as an example. First, the subsequence of 132 has 1, 3, 13, 2, 12, 32, 132. There is a conclusion that a number can be divisible by a certain number, then the sum of the numbers at all positions can be divisible by this number , ending in 2. The data is 2, 12, 32, 132. 12 can be regarded as $f["12"][mod]=f["12"][(mod+"2")\%mod]+f["1"]\left[mod\right]$ and so on.
3. Coding implementation

1      dp[1][1] = 1;
03     dp[2][0] = 1;
002    dp[3][2] = 1;

1  ⇒ 12  
13 ⇒ 132 ⇒ dp[3][(j+2)%3] = (dp[2][j] + dp[3][(j+2)%3])
2
3  ⇒ 32

for(int i = 1; i <= len; i++)
    for(int j = 0; j < 3; j++)
        for(int k = i+1; k <= len; k++)
            f[k][(s[k]+j-'0')%3] =(f[k][(s[k]+j-'0')%3]+ f[i][j])%mod;

#include <cstdio>
#include <cstring>；
using namespace std;
const int mod = 1e9 + 7;
char s[55];
int f[55][3];
int main()
{
    
    
    scanf("%s",s+1);
    int len = strlen(s+1);
    for(int i = 1; i <= len; i++) f[i][(s[i]-'0')%3] = 1;
    
    for(int i = 1; i <= len; i++)
        for(int j = 0; j < 3; j++)
            for(int k = i+1; k <= len; k++)
                f[k][(s[k]+j-'0')%3] =(f[k][(s[k]+j-'0')%3]+ f[i][j])%mod;
       
    int ans = 0;
    for(int i = 1; i <= len; i++) ans = (ans + f[i][0])%mod;
    printf("%d\n",ans);
    return 0;
}