牛客练习赛15 C.吉姆的奇思妙想(化简二分+溢出处理)

传送门

E i = a i ∗ ∑ 1 < = j < = n & & d e g j < = s ( d e g j 2 ∗ f r e j ) + b i ∗ ∑ 1 < = j < = n & & d e g j > s M ∗ f r e j = ∑ 1 < = j < = n & & d e g j < = s ( ( a i ∗ d e g j 2 − b i ∗ M ) ∗ f r e j ) + ∑ 1 < = j < = n b i ∗ M ∗ f r e j E_i=a_i*\sum\limits_{1<=j<=n\&\&deg_j<=s}(deg_j^2*fre_j)+b_i*\sum\limits_{1<=j<=n\&\&deg_j>s}M*fre_j\\ =\sum\limits_{1<=j<=n\&\&deg_j<=s}((a_i*deg_j^2-b_i*M)*fre_j)+\sum\limits_{1<=j<=n}b_i*M*fre_j\\ Ei​=ai​∗1<=j<=n&&degj​<=s∑​(degj2​∗frej​)+bi​∗1<=j<=n&&degj​>s∑​M∗frej​=1<=j<=n&&degj​<=s∑​((ai​∗degj2​−bi​∗M)∗frej​)+1<=j<=n∑​bi​∗M∗frej​

注意到 ∑ 1 < = j < = n b i ∗ M ∗ f r e j \sum\limits_{1<=j<=n}b_i*M*fre_j 1<=j<=n∑​bi​∗M∗frej​其实是个常数

而 a i ∗ d e g j 2 − b i ∗ M a_i*deg_j^2-b_i*M ai​∗degj2​−bi​∗M随时 j j j的增大而增大

所以只需要把所有令 a i ∗ d e g j 2 − b i ∗ M a_i*deg_j^2-b_i*M ai​∗degj2​−bi​∗M小于零的项找出来即可

找出这个最大的 j j j即可

话是这么说,但做的过程中千万小心爆掉long long

我用的是unsigned long long

所以判断负数只是比较大小关系,这样比较保险

非常坑

#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ll;
const ll maxn = 2e5+10;
ll m,l,deg[maxn],freq[maxn],pref[maxn],pred[maxn],sqrd[maxn];
int main()
{
	scanf("%llu%llu",&m,&l);
	for(int i=1;i<=l;i++)
	{
		scanf("%llu%llu",&deg[i],&freq[i] );
		pred[i] = pred[i-1]+deg[i]*deg[i]*freq[i];
		sqrd[i] = deg[i]*deg[i];
		pref[i] = pref[i-1]+freq[i];
	}
	int q; cin >> q;
	while( q-- )
	{
		ll a,b; cin >> a >> b;
		ll L = 1, R = l, ans = 0;
		while( R>=L )
		{
			ll mid = L+R>>1;
			if( a*sqrd[mid]<b*m )	L = mid+1,ans = mid;
			else	R = mid-1;
		}
		cout << a*pred[ans]+b*( pref[l]-pref[ans] )*m << endl;
	}
}

标签：化简,练习赛,pred,ll,long,奇思妙想,maxn,freq,deg
来源： https://blog.csdn.net/jziwjxjd/article/details/111146251