diff --git a/src/solution/mod.rs b/src/solution/mod.rs
index 2d39d64..abf3f89 100644
--- a/src/solution/mod.rs
+++ b/src/solution/mod.rs
@@ -7,3 +7,4 @@ mod s0003_longest_substring_without_repeating_characters;
 mod s0162_find_peak_element;
 mod s2828_check_if_a_string_is_an_acronym_of_words;
 mod s0052_n_queens_ii;
+mod s1276_number_of_burgers_with_no_waste_of_ingredients;
diff --git a/src/solution/s1276_number_of_burgers_with_no_waste_of_ingredients.rs b/src/solution/s1276_number_of_burgers_with_no_waste_of_ingredients.rs
new file mode 100644
index 0000000..84aef71
--- /dev/null
+++ b/src/solution/s1276_number_of_burgers_with_no_waste_of_ingredients.rs
@@ -0,0 +1,91 @@
+/**
+ * [1276] Number of Burgers with No Waste of Ingredients
+ *
+ * Given two integers tomatoSlices and cheeseSlices. The ingredients of different burgers are as follows:
+ * 
+ * 	Jumbo Burger: 4 tomato slices and 1 cheese slice.
+ * 	Small Burger: 2 Tomato slices and 1 cheese slice.
+ * 
+ * Return [total_jumbo, total_small] so that the number of remaining tomatoSlices equal to 0 and the number of remaining cheeseSlices equal to 0. If it is not possible to make the remaining tomatoSlices and cheeseSlices equal to 0 return [].
+ *  
+ * <strong class="example">Example 1:
+ * 
+ * Input: tomatoSlices = 16, cheeseSlices = 7
+ * Output: [1,6]
+ * Explantion: To make one jumbo burger and 6 small burgers we need 4*1 + 2*6 = 16 tomato and 1 + 6 = 7 cheese.
+ * There will be no remaining ingredients.
+ * 
+ * <strong class="example">Example 2:
+ * 
+ * Input: tomatoSlices = 17, cheeseSlices = 4
+ * Output: []
+ * Explantion: There will be no way to use all ingredients to make small and jumbo burgers.
+ * 
+ * <strong class="example">Example 3:
+ * 
+ * Input: tomatoSlices = 4, cheeseSlices = 17
+ * Output: []
+ * Explantion: Making 1 jumbo burger there will be 16 cheese remaining and making 2 small burgers there will be 15 cheese remaining.
+ * 
+ *  
+ * Constraints:
+ * 
+ * 	0 <= tomatoSlices, cheeseSlices <= 10^7
+ * 
+ */
+pub struct Solution {}
+
+// problem: https://leetcode.cn/problems/number-of-burgers-with-no-waste-of-ingredients/
+// discuss: https://leetcode.cn/problems/number-of-burgers-with-no-waste-of-ingredients/discuss/?currentPage=1&orderBy=most_votes&query=
+
+// submission codes start here
+
+impl Solution {
+    pub fn num_of_burgers(tomato_slices: i32, cheese_slices: i32) -> Vec<i32> {
+        let (mut tomato, mut cheese) = (tomato_slices as f32,
+                                    cheese_slices as f32);
+        if tomato == 0f32 && cheese == 0f32 {
+            return vec![0, 0]
+        }
+
+        if tomato == 0f32 ||
+            tomato / cheese < 2f32 || tomato / cheese > 4f32 {
+            return vec![]
+        }
+
+        let mut big_count = 0;
+
+        while tomato / cheese != 2f32 {
+            tomato -= 4f32;
+            cheese -= 1f32;
+            big_count += 1;
+
+            if tomato == 0f32 && cheese == 0f32 {
+                break;
+            }
+
+            if tomato * cheese == 0f32 {
+                return vec![]
+            }
+        }
+        
+        vec![big_count, cheese as i32]
+    }
+}
+
+// submission codes end
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    #[test]
+    fn test_1276() {
+        let empty_array : Vec<i32> = Vec::new();
+        assert_eq!(vec![1, 6], Solution::num_of_burgers(16, 7));
+        assert_eq!(empty_array, Solution::num_of_burgers(17, 4));
+        assert_eq!(empty_array, Solution::num_of_burgers(4, 17));
+        assert_eq!(vec![0, 0], Solution::num_of_burgers(0, 0));
+        assert_eq!(vec![0, 1], Solution::num_of_burgers(2, 1));
+    }
+}