AcWing 4956. 冶炼金属
原题链接
简单
二分 32ms
#include <bits/stdc++.h>
using namespace std;
const int N = 1e4 + 10;
int n;
int A[N], B[N];
bool check_max(int v) {
for (int i = 1; i <= n; i++)
if ((A[i] / v) < B[i]) return false;
return true;
}
bool check_min(int v) {
for (int i = 1; i <= n; i++)
if ((A[i] / v) > B[i]) return false;
return true;
}
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) scanf("%d%d", &A[i], &B[i]);
int l = 1, r = 1e9 + 10;
while (l < r) {
int mid = l + r >> 1;
if (check_min(mid)) r = mid;
else l = mid + 1;
}
printf("%d ", l);
l = 1, r = 1e9 + 10;
while (l < r) {
int mid = l + r + 1 >> 1;
if (check_max(mid)) l = mid;
else r = mid - 1;
}
printf("%d", l);
return 0;
}