CF670D2 Magic Powder

这题需要用到二分法，由于输入样例1<=k<=10^9，则最大的结果肯定小于等于2*10^9，因此我们在1~2*10^9这个范围内进行二次查找：

#include <bits/stdc++.h>
using namespace std;
int a[100010];
int b[100010];
int n, k, result;
void search(long long left, long long right)
{
long long mid = (left + right) / 2;
long long cnt = 0;
for (int i = 1; i <= n; i++)
{
if (mid * a[i] > b[i])
{
cnt += mid * a[i] - b[i];
}
if (cnt > k) break;
}
if (cnt > k)
{
if (left == right)
{
result = left - 1;
return ;
}
search(left, mid);
}
else if (cnt < k)
{
if (left == right)
{
result = left + 1;
return;
}
search(mid + 1, right);
}
else
{
result = mid;
return ;
}
}
int main()
{
cin >> n >> k;
for (int i = 1; i <= n; i++)
{
cin >> a[i];
}
for (int i = 1; i <= n; i++)
{
cin >> b[i];
}
search(1, 2000000000);
cout << result << endl;
return 0;
}