dpM - Candies
28 Jan 2021 — Tags: None
Click to show code.
using namespace std;
const int NMAX = 1e2 + 11;
const int KMAX = 1e5 + 11;
const int MOD = 1e9 + 7;
int n, k, a[NMAX], acc[KMAX], dp[NMAX][KMAX];
int solve(void)
{
dp[0][0] = 1;
for (int s = 1; s <= k; ++s)
dp[0][s] = 0;
for (int i = 1; i <= n; ++i)
{
acc[0] = dp[i - 1][0];
for (int s = 1; s <= k; ++s)
acc[s] = (acc[s - 1] + dp[i - 1][s]) % MOD;
for (int s = 0; s <= k; ++s)
dp[i][s] =
(acc[s] - (s - a[i] - 1 >= 0 ? acc[s - a[i] - 1] : 0) + MOD) %
MOD;
}
return dp[n][k];
}
int main(void)
{
cin >> n >> k;
for (int i = 1; i <= n; ++i)
cin >> a[i];
cout << solve() << endl;
return 0;
}