2435. 矩阵中和能被 K 整除的路径

src/main.rs

@@ -1,50 +1,30 @@
-use crate::arr::make_string_arr;
-use std::cmp::max;
+use crate::arr::make_matrix;
 
 struct Solution;
 mod arr;
 impl Solution {
-    pub fn find_max_form(strs: Vec<String>, m: i32, n: i32) -> i32 {
-        let mut dp = vec![vec![0; m as usize + 1]; n as usize + 1];
-        fn get_count(str: &String) -> (i32, i32) {
-            let (mut zeros, mut ones) = (0, 0);
-            for c in str.chars() {
-                if c == '0' {
-                    zeros += 1;
-                } else {
-                    ones += 1;
+    pub fn number_of_paths(grid: Vec<Vec<i32>>, k: i32) -> i32 {
+        let mut arr = vec![vec![vec![0; k as usize]; grid[0].len()]; grid.len()];
+        let r#mod = 1e9 as i32 + 7;
+        arr[0][0][(grid[0][0] % k) as usize] = 1;
+        for i in 0..grid.len() {
+            for j in 0..grid[i].len() {
+                if i < grid.len() - 1 {
+                    for l in 0..k {
+                        arr[i + 1][j][((l + grid[i + 1][j]) % k) as usize] = (arr[i + 1][j][((l + grid[i + 1][j]) % k) as usize] + arr[i][j][l as usize]) % r#mod;
+                    }
                 }
-            }
-            (zeros, ones)
-        }
-        let mut strs = strs
-            .iter()
-            .map(|x| get_count(x))
-            .collect::<Vec<(i32, i32)>>();
-        // n1 m0
-        // 外1内0
-
-        for i in 0..strs.len() {
-            let (zeros, ones) = strs[i];
-            for i in (ones..=n).rev() {
-                for k in (zeros..=m).rev() {
-                    dp[i as usize][k as usize] = dp[i as usize][k as usize]
-                        .max(dp[(i - ones) as usize][(k - zeros) as usize] + 1);
+                if j < grid[0].len() - 1 {
+                    for l in 0..k {
+                        arr[i][j + 1][((l + grid[i][j + 1]) % k) as usize] = (arr[i][j + 1][((l + grid[i][j + 1]) % k) as usize] + arr[i][j][l as usize]) % r#mod;
+                    }
                 }
             }
         }
-        let mut ans = 0;
-        dp.iter()
-            .for_each(|x| x.iter().for_each(|y| ans = max(ans, *y)));
-        ans
+        arr[grid.len() - 1][grid[0].len() - 1][0]
     }
 }
-// dp?
 fn main() {
-    let result = Solution::find_max_form(
-        make_string_arr("[\"10\", \"0001\", \"111001\", \"1\", \"0\"]"),
-        5,
-        3,
-    );
+    let result = Solution::number_of_paths(make_matrix("[[5,2,4],[3,0,5],[0,7,2]]"), 3);
     println!("{:?}", result);
 }