今天晚上的华为笔试第三题有做出来的吗？求思路

来个java版的 全ac的背包   public static void main(String[] args) { Scanner sc = new Scanner(System.in); int money = sc.nextInt(); List<Integer> list = new ArrayList<>(); while (sc.hasNext()) { list.add(sc.nextInt()); } int[] cost = new int[list.size() / 2]; int[] earn = new int[list.size() / 2]; int[] chosen = new int[list.size() / 2]; for (int i = 0; i < list.size(); ++i) { if (i < list.size() / 2) { cost[i] = list.get(i); chosen[i] = 0; } else { earn[i - list.size() / 2] = list.get(i); } } // d[i][j]表示前i个物品放入容量为j的背包的最大价值 // d[i][j] = max(d[i-1][j], d[i-1][j-weights[i]] + values[i]) int[][] d = new int[cost.length][money + 1]; for (int i = 0; i < d.length; ++i) { for (int j = 1; j < money + 1; ++j) { if (i == 0) { if (cost[i] <= j) { d[i][j] = cost[i]; } } else { if (j >= cost[i]) { d[i][j] = Math.max(d[i - 1][j], d[i - 1][j - cost[i]] + earn[i]); } else { d[i][j] = d[i - 1][j]; } } } } int ii = cost.length - 1, jj = money; while (ii >= 0) { if (ii > 0) { if (jj-cost[ii] >= 0 && d[ii][jj] == d[ii - 1][jj - cost[ii]] + earn[ii]) { chosen[ii] = 1; jj -= cost[ii]; } } else { if (cost[ii] <= jj) { chosen[ii] = 1; } } ii--; } boolean first = true; for (int i = 0; i < chosen.length; ++i) { if (chosen[i] == 1) { if (first) { System.out.print(i + 1); first = false; } else { System.out.print(" " + (i + 1)); } } } }

来个 Python AC的 coding=utf-8 def bag(n, c, w, v): res = [[-1 for j in range(c + 1)] for i in range(n + 1)] for j in range(c + 1): res[0][j] = 0 for i in range(1, n + 1): for j in range(1, c + 1): res[i][j] = res[i - 1][j] if j >= w[i - 1] and res[i][j] < res[i - 1][j - w[i - 1]] + v[i - 1]: res[i][j] = res[i - 1][j - w[i - 1]] + v[i - 1] return res def show(n, c, w, res): x = [0 for i in range(n)] j = c for i in range(n,0,-1): if res[i][j] > res[i - 1][j]: x[i - 1] = 1 j -= w[i - 1] for i in range(n): if x[i]==1: print(i+1), if name == 'main': c = int(raw_input()) w = raw_input() w = w.split() for i in range(len(w)): w[i] = int(w[i]) v = raw_input() v = v.split() for i in range(len(v)): v[i] = int(v[i]) n = len(v) res = bag(n, c, w, v) show(n, c, w, res)

深搜，第一遍深搜找最大价值，第二遍深搜找路径（本来想用背包做的，发现不会记录路径就换方法了）

直接递归就AC了

背包60，不知道为什么

背包60+1 dp数据我一步一步演算的,还是过不了神奇的测试用例

solve(); function solve () {     let         total = parseInt(readline()), // 总流量         consume = readline().split(/\s+/).map(s => Number(s)), // 每个app消耗的流量         coin = readline().split(/\s+/).map(s => Number(s)); // 每个app的金币数     let buffer = [];     for (let i = 0; i < consume.length + 1; i++) {         buffer[i] = [0];     }     for (let w = 0; w < total + 1; w++) {         buffer[0][w] = 0;     }     for (let w = 1; w < total + 1; w++) {         for (let i = 1; i < consume.length + 1; i++) {             if (consume[i - 1] > w) {                 buffer[i][w] = buffer[i - 1][w];             } else {                 buffer[i][w] = Math.max(buffer[i - 1][w], coin[i - 1] + buffer[i - 1][w - consume[i-1]]);             }         }     }     let result = [];     let w = total;     for (let i = consume.length; i > 0; i--) {         if (buffer[i][w] === buffer[i-1][w]) {             continue;         } else {             result.push(i);             w -= consume[i - 1];         }     }     print(result.reverse().join(' ')); } 背包的解法

01背包记录路径

典型背包问题，5分钟AC

背包问题输出背包序列。