腾讯二面上机题

哪个岗位的啊😂

dp(i)表示跳到第i个格子的最大体力，转移方程dp(i)=dp(i-x)-x+a(i)

    public static int MaxValue(int m, ArrayList<Integer> list) {         int start = 0;         int index = start;         for (; index < m; index++) {             if (index >= list.size()) {                 break;             }             if (list.get(index) > 0) {                 m += list.get(index);                 m -= (index - start);                 start = index;             }         }         if (index >= list.size()) {             return m;         } else {             return -1;         }     } 不知道对不对，没有oj能测试了

动态规划

广搜？