Aug 23, 2024 · Aug 8, 2024 · Aug 8, 2024 · Aug 9, 2024 · Aug 9, 2024 · Aug 23, 2024
diff --git a/problems/背包理论基础01背包-1.md b/problems/背包理论基础01背包-1.md
diff --git a/problems/背包理论基础01背包-2.md b/problems/背包理论基础01背包-2.md
Original file line number	Diff line number	Diff line change
Expand Up		@@ -437,6 +437,58 @@ print(dp[n - 1][bagweight])
		### Go

		```go
	package main

	import (
	"fmt"
	)

	func main() {
	var n, bagweight int // bagweight代表行李箱空间
	fmt.Scan(&n, &bagweight)

	weight := make([]int, n) // 存储每件物品所占空间
	value := make([]int, n) // 存储每件物品价值

	for i := 0; i < n; i++ {
	fmt.Scan(&weight[i])
	}
	for j := 0; j < n; j++ {
	fmt.Scan(&value[j])
	}
	// dp数组, dp[i][j]代表行李箱空间为j的情况下,从下标为[0, i]的物品里面任意取,能达到的最大价值
	dp := make([][]int, n)
	for i := range dp {
	dp[i] = make([]int, bagweight+1)
	}

	// 初始化, 因为需要用到dp[i - 1]的值
	// j < weight[0]已在上方被初始化为0
	// j >= weight[0]的值就初始化为value[0]
	for j := weight[0]; j <= bagweight; j++ {
	dp[0][j] = value[0]
	}

	for i := 1; i < n; i++ { // 遍历科研物品
	for j := 0; j <= bagweight; j++ { // 遍历行李箱容量
	if j < weight[i] {
	dp[i][j] = dp[i-1][j] // 如果装不下这个物品,那么就继承dp[i - 1][j]的值
	} else {
	dp[i][j] = max(dp[i-1][j], dp[i-1][j-weight[i]]+value[i])
	}
	}
	}

	fmt.Println(dp[n-1][bagweight])
	}

	func max(x, y int) int {
	if x > y {
	return x
	}
	return y
	}

		```

		### Javascript
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Original file line number	Diff line number	Diff line change
Expand Up		@@ -318,50 +318,45 @@ print(dp[n - 1][bagweight])
		package main

		import (
	"fmt"
	"fmt"
		)

		func main() {
	var n, bagweight int
	fmt.Scan(&n, &bagweight)

	weight := make([]int, n)
	value := make([]int, n)

	for i := 0; i < n; i++ {
	fmt.Scan(&weight[i])
	}
	for j := 0; j < n; j++ {
	fmt.Scan(&value[j])
	}

	dp := make([][]int, n)
	for i := range dp {
	dp[i] = make([]int, bagweight+1)
	}

	for j := weight[0]; j <= bagweight; j++ {
	dp[0][j] = value[0]
	}

	for i := 1; i < n; i++ {
	for j := 0; j <= bagweight; j++ {
	if j < weight[i] {
	dp[i][j] = dp[i-1][j]
	} else {
	dp[i][j] = max(dp[i-1][j], dp[i-1][j-weight[i]]+value[i])
	}
	}
	}

	fmt.Println(dp[n-1][bagweight])
	// 读取 M 和 N
	var M, N int
	fmt.Scan(&M, &N)

	costs := make([]int, M)
	values := make([]int, M)

	for i := 0; i < M; i++ {
	fmt.Scan(&costs[i])
	}
	for j := 0; j < M; j++ {
	fmt.Scan(&values[j])
	}

	// 创建一个动态规划数组dp,初始值为0
	dp := make([]int, N + 1)

	// 外层循环遍历每个类型的研究材料
	for i := 0; i < M; i++ {
	// 内层循环从 N 空间逐渐减少到当前研究材料所占空间
	for j := N; j >= costs[i]; j-- {
	// 考虑当前研究材料选择和不选择的情况,选择最大值
	dp[j] = max(dp[j], dp[j-costs[i]] + values[i])
	}
	}

	// 输出dp[N],即在给定 N 行李空间可以携带的研究材料最大价值
	fmt.Println(dp[N])
		}

	func max(a, b int) int {
	if a > b {
	return a
	}
	return b
	func max(x, y int) int {
	if x > y {
	return x
	}
	return y
		}

		```
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