CSES-solutions/CSES Problem Set/Array Description.cpp
Segcolt a82ebbc4da Add a lot of cses dp problems
I had some free time to just do them, they were pretty easy though.
2024-10-04 17:45:55 -03:00

120 lines
2.5 KiB
C++

/* Problem URL: https://cses.fi/problemset/task/1746 */
#include <bits/stdc++.h>
using namespace std;
#define V vector
#define rmin(a, b) a = min(a, b)
#define rmax(a, b) a = max(a, b)
#define rep(i, lim) for (size_t i = 0; i < (lim); i++)
#define nrep(i, s, lim) for (size_t i = s; i < (lim); i++)
#define repv(i, v) for (auto &i : (v))
#define fillv(v) for (auto &itr_ : (v)) { cin >> itr_; }
#define sortv(v) sort(v.begin(), v.end())
#define all(v) (v).begin(), (v).end()
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using vvvvi = vector<vvvi>;
using ll = long long;
using vl = vector<ll>;
using vvl = vector<vl>;
using vvvl = vector<vvl>;
using vvvvl = vector<vvvl>;
template<class v>
auto operator<<(ostream &os, const vector<v> &vec)->ostream& {
os << vec[0];
for (size_t i = 1; i < vec.size(); i++) {
os << ' ' << vec[i];
}
os << '\n';
return os;
}
template<class v>
auto operator>>(istream &is, vector<v> &vec)->istream& {
for (auto &i : vec) {
is >> i;
}
return is;
}
template<class v>
auto operator<<(ostream &os, const vector<vector<v>> &vec)->ostream& {
for (auto &i : vec) {
os << i[0];
for (size_t j = 1; j < i.size(); j++) {
os << ' ' << i[j];
}
os << '\n';
}
return os;
}
template<class v>
auto operator>>(istream &is, vector<vector<v>> &vec)->istream& {
for (auto &i : vec) {
for (auto &j : i) {
is >> j;
}
}
return is;
}
const ll mod = 1e9 + 7;
int main()
{
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n, m;
cin >> n >> m;
vi nums(n);
cin >> nums;
vvl dp(n, vl(m + 1, 0));
if (nums[0] > 0) {
dp[0][nums[0]] = 1;
} else {
fill(dp[0].begin(), dp[0].end(), 1);
dp[0][0] = 0;
}
nrep(i, 1, n) {
if (nums[i] > 0) {
ll ans = dp[i - 1][nums[i]];
if (nums[i] < m) {
ans += dp[i - 1][nums[i] + 1];
}
if (nums[i] > 1) {
ans += dp[i - 1][nums[i] - 1];
}
dp[i][nums[i]] = ans % mod;
continue;
}
for (size_t j = 1; j <= m; j++) {
dp[i][j] += dp[i - 1][j];
if (j > 1) {
dp[i][j] += dp[i - 1][j - 1];
}
if (j < m) {
dp[i][j] += dp[i - 1][j + 1];
}
dp[i][j] %= mod;
}
}
cout << accumulate(dp[n - 1].begin(), dp[n - 1].end(), 0LL) % mod << '\n';
}