UPC 抢金块（普通的状态转移）

此题的初始状态是1，终末状态位于n，而状态转移范围为s~t，所以对于每一状态，只需枚举其合法的转移方式并取最大值即可。

#include<pch.h>
#include <iostream>
#include <cstdio>
#include <bits/stdc++.h>
#include <map>
#include <algorithm>
#include <stack>
#include <iomanip>
#include <cstring>
#include <cmath>
#define DETERMINATION main
#define lldin(a) scanf_s("%lld", &a)
#define println(a) printf("%lld\n", a)
#define reset(a, b) memset(a, b, sizeof(a))
const int INF = 0x3f3f3f3f;
using namespace std;
const double PI = acos(-1);
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int mod = 1000000007;
const int tool_const = 19991126;
const int tool_const2 = 2000;
inline ll lldcin()
{
	ll tmp = 0, si = 1;
	char c;
	c = getchar();
	while (c > '9' || c < '0')
	{
		if (c == '-')
			si = -1;
		c = getchar();
	}
	while (c >= '0' && c <= '9')
	{
		tmp = tmp * 10 + c - '0';
		c = getchar();
	}
	return si * tmp;
}
///Untersee Boot IXD2(1942)
/**Although there will be many obstructs ahead,
the desire for victory still fills you with determination..**/
/**Last Remote**/
ll a[5000],dp[5000];
int DETERMINATION()
{
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	ll n;
	cin >> n;
	ll s, t;
	cin >> s >> t;
	for (int i = 1; i <= n; i++)
		cin >> a[i];
	dp[1] = a[1];
	for (int i = s+1; i <= n; i++)
		for (int j = s; j <= t; j++)
				dp[i] = max(dp[i], dp[i - j] + a[i]);
//第二个for循环就是dp[i]=max(dp[i-s]+a[i],dp[i-s-1]+a[i],...,dp[i-t]+a[i])的另一个表现形式
	cout << dp[n] << endl;
	return 0;
}