Submission
Status:
xP-PPx--P-PPPPPP
Score: 100
User: Nagornz
Problemset: Chocolate
Language: cpp
Time: 0.048 second
Submitted On: 2025-03-14 22:06:21
#include <bits/stdc++.h>
#define int long long
#define double long double
#define pii pair <int,int>
#define tiii tuple <int, int, int>
#define f first
#define s second
#define all(x) x.begin(), x.end()
#define ub(a, b) upper_bound(a.begin(), a.end(), b) - a.begin();
#define lb(a, b) lower_bound(a.begin(), a.end(), b) - a.begin();
#define ve vector
#define graph(a, n) vector <int> a[n];
#define wgraph(a, n) vector <pii> a[n];
#define emb emplace_back
#define em emplace
#define ins insert
#define er erase
#define iShowSpeed cin.tie(NULL)->sync_with_stdio(false)
using namespace std;
template <typename T>
using greater_priority_queue = priority_queue<T, vector<T>, greater<T>>;
const int mod = 1e9 + 7;
const int inf = 1e18;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
int power(int a, int n){
int ans = 1;
while (n) {
if (n % 2) ans *= a;
a *= a;
n /= 2;
}
return ans;
}
int32_t main(){
iShowSpeed;
int n, k, c; cin >> n >> k >> c;
vector <int> a(n + 1); for (int i = 1; i <= n; i++) cin >> a[i];
int ll, rr; cin >> ll >> rr;
vector <vector <int>> pref(n + 1, vector <int> (n + 1));
for (int i = 1; i <= n; i++) {
for (int j = i; j <= n; j++) {
pref[i][j] = pref[i][j - 1] + a[j] * power(j - i, c);
// cout << i << " -> " << j << ": " << pref[i][j] << "\n";
}
}
vector <int> dp(n + 1), prev(n + 1);
for (int i = 1; i <= n; i++) {
dp[i] = (pref[1][i] >= ll && pref[1][i] <= rr);
}
for (int i = 2; i <= k; i++) {
prev = dp;
for (int j = 1; j <= n; j++) {
prev[j] += prev[j - 1];
}
for (int j = i; j <= n; j++) {
int l = 1, r = j;
int lidx = 0;
while (l <= r) {
int mid = (l + r) / 2;
if (pref[mid][j] <= rr) {
r = mid - 1;
lidx = mid;
}
else {
l = mid + 1;
}
}
l = 1, r = j;
int ridx = 0;
while (l <= r) {
int mid = (l + r) / 2;
if (pref[mid][j] >= ll) {
l = mid + 1;
ridx = mid;
}
else {
r = mid - 1;
}
}
dp[j] = (ridx - 1 < 0 ? 0 : prev[ridx - 1]) - (lidx == 0 ? 1e14 : (lidx == 1 ? 0 : prev[lidx - 2]));
dp[j] = (dp[j] + mod) % mod;
}
}
cout << max(0ll, dp[n]);
}