Codeforces 444B DZY Loves FFT (probability)

Codeforces 444B DZY Loves FFT (probability)

Question connection: Codeforces 444B DZY Loves FFT

Generate arrays a and B based on the algorithm of the question, and then calculate the maximum value of a [I] * B [l-I] for each length of l.

Solution: probability problem. If there is a number of the first 30 values in the enumeration, you can directly output the answer. If not, you can find the maximum value at the position where the B array is 1.

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 1e5;const int s = 30;int n, d, x, rec[maxn+5];int np, pos[maxn+5];int a[maxn+5], b[maxn+5];int get_next () {    return x = (37LL * x + 10007) % 1000000007;}void init_table () {    for (int i = 0; i < n; i++)        a[i] = i + 1;    for (int i = 0; i < n; i++)        swap(a[i], a[get_next() % (i+1)]);    for (int i = 0; i < n; i++)        b[i] = (i < d ? 1 : 0);    for (int i = 0; i < n; i++)        swap(b[i], b[get_next() % (i+1)]);}void init_pos () {    np = 0;    for (int i = 0; i < n; i++) {        if (b[i])            pos[np++] = i;    }    for (int i = 0; i < n; i++)        rec[a[i]] = i;}int main () {    scanf("%d%d%d", &n, &d, &x);    init_table();    init_pos();    for (int i = 0; i < n; i++) {        bool flag = true;        for (int j = n; j >= n-s; j--) {            if (rec[j] <= i && b[i-rec[j]]) {                flag = false;                printf("%d\n", j);                break;            }        }        if (flag) {            int ans = 0;            for (int j = 0; j < np && pos[j] <= i; j++) {                ans = max(ans, a[i-pos[j]]);            }            printf("%d\n", ans);        }    }    return 0;}