# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
26609 | model_code | Lollipop (POI11_liz) | C11 | 633 ms | 17716 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
/*************************************************************************
* *
* XVIII Olimpiada Informatyczna *
* *
* Zadanie: Lizak *
* Autor: Mateusz Baranowski *
* Zlozonosc czasowa: O(n + m) *
* pamieciowa: O(n) *
* Opis: Rozwiazanie bledne *
* Nie uwzglednia, ze segment waniliowy moze byc *
* blizej "prawego konca" *
* *
*************************************************************************/
#include <stdio.h>
#define MAX_N 1000000
#define MAX_K 2000000
int n, m; /* ilosc segmentow lizaka i ilosc rozwazanych cen */
int wa[MAX_K + 1], wb[MAX_K + 1]; /* [wa[j], wb[j]] to przedzial o koszcie j */
int i, l, sum = 0;
int max[2]; /* najwieksza cena parzysta i nieparzysta */
char s[MAX_N + 1]; /* opis lizaka */
/* funkcja obliczajaca wa[x], wb[x] dla (x <= sum) i ((sum + x) mod 2 == 0) */
void oblicz_przedzialy() {
int sum_tmp = sum;
wa[sum] = i + 1;
wb[sum] = l + 1;
while (sum_tmp > 2) {
while ((sum_tmp > 2) && (s[i] == 'T')) {
sum_tmp -= 2;
wa[sum_tmp] = ++i + 1;
wb[sum_tmp] = l + 1;
}
while ((sum_tmp > 2) && (s[l] == 'T')) {
sum_tmp -= 2;
wa[sum_tmp] = i + 1;
wb[sum_tmp] = --l + 1;
}
if (sum_tmp > 2) {
sum_tmp -= 2;
wa[sum_tmp] = ++i + 1;
wb[sum_tmp] = --l + 1;
}
}
}
int main() {
scanf("%d %d", &n, &m);
scanf("%s", s);
for (l = 0; l < n; ++l)
if (s[l] == 'T')
sum += 2;
else
sum += 1;
i = 0;
l = n - 1;
max[sum % 2] = sum;
oblicz_przedzialy();
/* szukamy pierwszego 'W' od poczatku lizaka */
i = 0;
while ((i < n) && (s[i] == 'T'))
++i;
if (i == n) /* brak waniliowego */
max[1 - (sum % 2)] = 0;
else {
sum -= 2 * i + 1;
max[sum % 2] = sum;
i += 1;
l = n - 1;
oblicz_przedzialy();
}
/* odpowiadamy na zapytania korzystajac z tablic wa[] i wb[] */
for (i = 0; i < m; ++i) {
scanf("%d", &sum);
if (max[sum % 2] < sum)
printf("NIE\n");
else
printf("%d %d\n", wa[sum], wb[sum]);
}
return 0;
}
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