题目传送门
sol:二分答案$K$，算大于$K$的乘积有多少个。关键在于怎么算这个个数，官方题解上给出的复杂度是$O(nlogn)$，那么计算个数的复杂度是$O(n)$的。感觉写着有点困难，自己写了一个复杂度是$O(nlog^{2}n)$，也够AC了。有正有负，控制边界有点难度。
- 二分答案
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int MAXN = 1e5 + ;
int a[MAXN], b[MAXN];
int n, m; LL k;
LL check(LL mid) {
LL sum = ;
for (int i = ; i <= n; i++) {
if (a[i] == ) {
if (mid < ) sum += m;
continue;
}
if (a[i] > ) {
if (mid >= ) {
sum += b + + m - upper_bound(b + , b + + m, mid / a[i]);
} else {
sum += b + + m - lower_bound(b + , b + + m, (mid + ) / a[i]);
}
} else {
if (mid >= ) {
sum += lower_bound(b + , b + + m, mid / a[i]) - - (b + - );
} else {
sum += upper_bound(b + , b + + m, (mid + ) / a[i]) - - (b + - );
}
}
}
return sum;
}
int main() {
scanf("%d%d%lld", &n, &m, &k);
for (int i = ; i <= n; i++) scanf("%d", &a[i]);
for (int i = ; i <= m; i++) scanf("%d", &b[i]);
sort(b + , b + + m);
LL l = -1e12 - , r = 1e12 + ;
while (l + < r) {
LL mid = l + r >> ;
if (check(mid) < k) r = mid;
else l = mid;
}
printf("%lld\n", r);
return ;
}