Stan Ford is a typical college graduate student, meaning that one of the most important things on his mind
is where his next meal will be. Fortune has smiled on him as he’s been invited to a multi-course barbecue
put on by some of the corporate sponsors of his research team, where each course lasts exactly one hour.
Stan is a bit of an analytical type and has determined that his eating pattern over a set of consecutive hours
is always very consistent. In the first hour, he can eat up to m calories (where m depends on factors such
as stress, bio-rhythms, position of the planets, etc.), but that amount goes down by a factor of two-thirds
each consecutive hour afterwards (always truncating in cases of fractions of a calorie). However, if he stops
eating for one hour, the next hour he can eat at the same rate as he did before he stopped. So, for example,
if m = 900 and he ate for five consecutive hours, the most he could eat each of those hours would be 900,
600, 400, 266 and 177 calories, respectively. If, however, he didn’t eat in the third hour, he could then eat
900, 600, 0, 600 and 400 calories in each of those hours. Furthermore, if Stan can refrain from eating for two
hours, then the hour after that he’s capable of eating m calories again. In the example above, if Stan didn’t
eat during the third and fourth hours, then he could consume 900, 600, 0, 0 and 900 calories.
Stan is waiting to hear what will be served each hour of the barbecue as he realizes that the menu will
determine when and how often he should refrain from eating. For example, if the barbecue lasts 5 hours and
the courses served each hour have calories 800, 700, 400, 300, 200 then the best strategy when m = 900 is
to eat every hour for a total consumption of 800 + 600 + 400 + 266 + 177 = 2 243 calories. If however, the
third course is reduced from 400 calories to 40 calories (some low-calorie celery dish), then the best strategy
is to not eat during the third hour — this results in a total consumption of 1 900 calories.
The prospect of all this upcoming food has got Stan so frazzled he can’t think straight. Given the number of
courses and the number of calories for each course, can you determine the maximum amount of calories Stan
can eat?
Input
Input starts with a line containing two positive integers n m (n ≤ 100, m ≤ 20 000) indicating the number
of courses and the number of calories Stan can eat in the first hour, respectively. The next line contains n
positive integers indicating the number of calories for each course.
Output
Display the maximum number of calories Stan can consume.
Sample Input 1 Sample Output 1
5 900
800 700 400 300 200
2243

考虑dp[ i ][ j ]表示准备吃第i 个食物，前面已经吃了 j 个食物的决策情况，然后枚举中途休息的情况即可
一道不错的dp题

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<string>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<sstream>
typedef long long ll;
using namespace std;
typedef unsigned long long int ull;
#define maxn 600005
#define ms(x) memset(x,0,sizeof(x))
#define Inf 0x7fffffff
#define inf 0x3f3f3f3f
const long long int mod = 1e9 + 7;
#define pi acos(-1.0)
#define pii pair<int,int>
#define eps 1e-7
#define pll pair<ll,ll>



ll quickpow(ll a, ll b) {
    ll ans = 1;
    a = a % mod;
    while (b > 0) {
        if (b % 2)ans = ans * a;
        b = b / 2;
        a = a * a;
    }
    return ans;
}

int gcd(int a, int b) {
    return b == 0 ? a : gcd(b, a%b);
}

int n, m;
int a[maxn];
int dp[200][200];
int b[200];
int main()
{
    ios::sync_with_stdio(false);
    cin >> n >> m;
    int i, j;
    for (i = 1; i <= n; i++) {
        cin >> a[i];
    }
    int maxx = -inf;
    b[0] = m;
    for (i = 1; i <= 200; i++)
        b[i] = b[i - 1] * 2 / 3;
    for (i = 1; i <= n; i++) {
        dp[i][0] = min(a[i], b[0]);
    }
    for (i = 1; i <= n; i++) {
        for (j = 0; j <= n; j++) {
            for (int k = 1; k <= n && k + i <= n&&k<=2; k++) {
                dp[i + k][j + 2 - k] = max(dp[i + k][j + 2 - k], dp[i][j] + min(a[i + k], b[j + 2 - k]));
            }
            dp[i + 3][0] = max(dp[i + 3][0], dp[i][j] + min(m, a[i + 3]));
        }
    }
    for (i = 0; i <= n; i++) {
        maxx = max(maxx, dp[n][i]);
    }
    cout << maxx << endl;
}