結果

問題 No.1816 MUL-DIV Game
ユーザー 👑 ygussanyygussany
提出日時 2022-01-21 21:29:33
言語 C
(gcc 13.3.0)
結果
AC  
実行時間 42 ms / 2,000 ms
コード長 3,479 bytes
コンパイル時間 1,387 ms
コンパイル使用メモリ 30,592 KB
実行使用メモリ 6,824 KB
最終ジャッジ日時 2024-11-25 22:41:41
合計ジャッジ時間 2,691 ms
ジャッジサーバーID
(参考情報)
judge4 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 27
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include <stdio.h>
typedef struct {
long long key;
int id;
} data;
typedef struct {
data obj[100001];
int size;
} min_max_heap;
long long get_min(min_max_heap* h)
{
return h->obj[0].key;
}
int get_argmin(min_max_heap* h)
{
return h->obj[0].id;
}
long long get_max(min_max_heap* h)
{
if (h->size == 1) return h->obj[0].key;
else return h->obj[1].key;
}
int get_argmax(min_max_heap* h)
{
if (h->size == 1) return h->obj[0].id;
else return h->obj[1].id;
}
void push(min_max_heap* h, data x)
{
int i = (h->size)++;
data tmp;
if ((i & 1) != 0 && x.key < h->obj[i^1].key) {
h->obj[i] = h->obj[i^1];
h->obj[i^1] = x;
i ^= 1;
} else h->obj[i] = x;
int j, k = ((i >> 1) + 1) >> 1;
if (k == 0) return;
j = (k - 1) << 1;
if (h->obj[i].key < h->obj[j].key) {
while (k > 0) {
if (h->obj[i].key < h->obj[j].key) {
tmp = h->obj[j];
h->obj[j] = h->obj[i];
h->obj[i] = tmp;
k >>= 1;
i = j;
j = (k - 1) << 1;
} else break;
}
} else if (h->obj[i].key > h->obj[j^1].key) {
j ^= 1;
while (k > 0) {
if (h->obj[i].key > h->obj[j].key) {
tmp = h->obj[j];
h->obj[j] = h->obj[i];
h->obj[i] = tmp;
k >>= 1;
i = j;
j = ((k - 1) << 1) ^ 1;
} else break;
}
}
}
data pop_min(min_max_heap* h)
{
int i = 0, j = 2, k = 2;
data output = h->obj[0], tmp;
h->obj[0] = h->obj[--(h->size)];
while (j < h->size) {
if (j + 2 < h->size && h->obj[j+2].key < h->obj[j].key) {
j += 2;
k++;
}
if (h->obj[i].key > h->obj[j].key) {
tmp = h->obj[j];
h->obj[j] = h->obj[i];
h->obj[i] = tmp;
k <<= 1;
i = j;
j = (k - 1) << 1;
} else break;
}
if (j < h->size) return output;
if (i < h->size - 1 && h->obj[i].key > h->obj[i^1].key) {
tmp = h->obj[i^1];
h->obj[i^1] = h->obj[i];
h->obj[i] = tmp;
i ^= 1;
}
k = ((i >> 1) + 1) >> 1;
j = ((k - 1) << 1) ^ 1;
while (k > 0) {
if (h->obj[i].key > h->obj[j].key) {
tmp = h->obj[j];
h->obj[j] = h->obj[i];
h->obj[i] = tmp;
k >>= 1;
i = j;
j = ((k - 1) << 1) ^ 1;
} else break;
}
return output;
}
data pop_max(min_max_heap* h)
{
if (h->size == 1) return h->obj[--(h->size)];
int i = 1, j = 2, k = 2;
data output = h->obj[1], tmp;
h->obj[1] = h->obj[--(h->size)];
while (j < h->size) {
if (j + 1 < h->size) {
j++;
if (j + 2 < h->size && h->obj[j+2].key > h->obj[j].key) {
j += 2;
k++;
} else if (j + 1 < h->size && h->obj[j+1].key > h->obj[j].key) {
j++;
k++;
}
}
if (h->obj[i].key < h->obj[j].key) {
tmp = h->obj[j];
h->obj[j] = h->obj[i];
h->obj[i] = tmp;
k <<= 1;
i = j;
j = (k - 1) << 1;
} else break;
}
if (j < h->size) return output;
i = (i >> 1) << 1;
if (i < h->size - 1 && h->obj[i].key > h->obj[i^1].key) {
tmp = h->obj[i^1];
h->obj[i^1] = h->obj[i];
h->obj[i] = tmp;
}
k = ((i >> 1) + 1) >> 1;
j = (k - 1) << 1;
while (k > 0) {
if (h->obj[i].key < h->obj[j].key) {
tmp = h->obj[j];
h->obj[j] = h->obj[i];
h->obj[i] = tmp;
k >>= 1;
i = j;
j = (k - 1) << 1;
} else break;
}
return output;
}
int main()
{
int i, N;
data d;
min_max_heap h;
h.size = 0;
scanf("%d", &N);
for (i = 1; i <= N; i++) {
scanf("%d", &(d.key));
push(&h, d);
}
for (i = 1; i < N; i++) {
if (i % 2 == 0) {
pop_max(&h);
pop_max(&h);
d.key = 1;
push(&h, d);
} else {
d = pop_min(&h);
d.key *= pop_min(&h).key;
push(&h, d);
}
}
printf("%lld\n", get_max(&h));
fflush(stdout);
return 0;
}
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