传送门：点击打开链接
题意：有n个箱子排成一排，有m个炸弹，位置告诉你，现在炸弹的左边伤害和右边伤害可以自己控制，要求 每个炸弹炸的箱子数的累乘，输出答案取log2并乘以1e6
思路：直接2for xjb搞就行了，大概就是某个区间里刚好只有一个炸弹时，就是满足的，然后就从前面往后面更新一下
#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <stack>
#include <queue>
#include <cstdio>
#include <cctype>
#include <string>
#include <vector>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#include <functional>
#define fuck(x) cout<<"["<<x<<"]"
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w+",stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;const int MX = 2e3 + 5;
const int W = 1e6;int A[MX];
double dp[MX];int main() {int T, n, m; //FIN;scanf("%d", &T);while(T--) {scanf("%d%d", &n, &m);memset(A, 0, sizeof(A));memset(dp, 0, sizeof(dp));for(int i = 1; i <= m; i++) {int t; scanf("%d", &t);t++; A[t] = 1;}for(int i = 2; i <= n; i++) A[i] += A[i - 1];for(int i = 1; i <= n; i++) {for(int j = 0; j < i; j++) {if(A[i] - A[j] == 1) {dp[i] = max(dp[i], dp[j] + log2(1.0 * i - j));}}}printf("%.0f\n", floor(dp[n] * W));}
}