이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
//khodaya khodet komak kon
#include <bits/stdc++.h>
#define int long long
#define F first
#define S second
#define pb push_back
#define all(x) x.begin(), x.end()
#pragma GCC optimize ("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize ("-O2")
//9:00
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
const int N = 10000 + 10;
const ll MOD = 1000000000 + 7;
const ll INF = 1000000010;
const ll LOG = 25;
int dp[2][N][2], n, a[N];
int32_t main(){
ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
cin >> n;
for (int i = 1; i <= n; i++) cin >> a[i];
bool f = 1;
dp[0][0][0] = 1;
int mx = 0;
for (int i = 1; i <= n; i++){
int nmx = max(mx, a[i]);
memset(dp[f], 0, sizeof dp[f]);
dp[f][nmx][0] = dp[f ^ 1][mx][0];
//cout << dp[f][nmx][0] << '\n';
for (int j = 1; j <= i; j++){
dp[f][j][1] = (dp[f ^ 1][j - 1][1] + dp[f ^ 1][j][1] * 1ll * j % MOD) % MOD;
}
dp[f][mx][1] = (dp[f][mx][1] + dp[f ^ 1][mx][0] * 1ll * min(mx, a[i] - 1) % MOD) % MOD;
mx = nmx;
f ^= 1;
}
ll res = 0;
for (int i = 0; i <= n; i++){
res += dp[f ^ 1][i][1];
// cout << i << ' ' << dp[f ^ 1][i][1] << '\n';
res %= MOD;
}
res = (res + 1) % MOD;
cout << res;
return 0;
}
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