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// 2327. Number of People Aware of a Secret
#include "leetcode.h"
/*
* on day 1, one person discovers decret. you are given an integer 'delay'
* which means that each person will share the secret with a new person every
* day starting from 'delay' days after discovering the secret. you are also
* given an integer 'forget', which means that each person will forget the
* secret 'forget' days after discovering it. a person cannot share the secret
* on the same day they forgot it, or any day afterwards. given an integer 'n',
* return the number of people who know the secret at the end of day 'n'. since
* the answer may be very large, return it modulo 10^9+7.
*/
int peopleAwareOfSecret(int n, int delay, int forget) {
long *dp = (long *)malloc((n + 1) * sizeof(int));
dp[1] = 1;
int share = 0, mod = 1e9 + 7, ans = 0;
for (int i = 2; i <= n; ++i)
dp[i] = share = (share + dp[(int)fmax(i - delay, 0)] -
dp[(int)fmax(i - forget, 0)] + mod) %
mod;
for (int i = n - forget + 1; i <= n; ++i)
ans = (ans + dp[i]) % mod;
free(dp);
return ans;
}
int main() {
printf("%d\n", peopleAwareOfSecret(6, 2, 4)); // expect: 5
printf("%d\n", peopleAwareOfSecret(4, 1, 3)); // expect: 6
}
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