快快变大(区间合并)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;   
typedef vector<ll> Vll;               
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;                        

const ll mod = 1e6 + 3;
//const ll mod = 998244353;
const double pi  = acos(-1.0);

inline ll qmi (ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans = ans * a%mod;
        a = a * a %mod;
        b >>= 1;
    }
    return ans;
}
inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
    return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
    return (a + b) %mod;
}
inline ll inv (ll a) {
    return qmi(a, mod - 2);
}
int n;
ll dp[1000][1000];
ll sum[1000];
int a[1000]; 

ll cal (ll a, ll b) {
    return sum[b] * inv(sum[a - 1]) %mod;
}
ll pow2 (ll a) {
    return a * a;
}
void solve() {
    cin >> n;
    for (int i = 1; i <= n; i ++)
        cin >> a[i];
    sum[0] = 1;
    for (int i = 1;  i<= n; i ++) {
        sum[i] = sum[i - 1] * a[i]%mod;
    }
    
    for (int len = 2; len <= n; len ++) {
        for (int l = 1; l + len - 1 <= n; l ++) {
            int r = l + len - 1;
            for (int mid = l; mid < r; mid ++) {
                
                dp[l][r] = max (dp[l][r],
                dp[l][mid] + dp[mid + 1][r] + 
                pow2(cal(l, mid) - cal(mid + 1, r)));
            }
        }
    }
    cout << dp[1][n] << endl;
}
int main () {
    // ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}

题意:类似与石子合并

题解:通过区间合并可以解决
dp[i][j]表示将i,j的石子合并得到的最优解

我一开始是用贪心的角度去想，但是没有发现规律。甚至没有发现这是个dp。这个问题的结构特征:每次选择相邻的石子合并.有明显的方向