bzoj 4590 自动刷题机

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二分答案.

  • 不难发现 $n$ 增大,切题数不会增多, $n$ 减小,切题数不会减少.
  • 于是分别二分 $n$ 的最小值与最大值,检验直接模拟操作就好了.
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#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
inline int read()
{
	int out=0,fh=1;
	char jp=getchar();
	while ((jp>'9'||jp<'0')&&jp!='-')
		jp=getchar();
	if (jp=='-')
		fh=-1,jp=getchar();
	while (jp>='0'&&jp<='9')
		out=out*10+jp-'0',jp=getchar();
	return out*fh;
}
const int MAXN=1e5+10;
int n,k,x[MAXN];
bool checkmin(ll mid)
{
	int tot=0;
	ll len=0;
	for(int i=1;i<=n;++i)
	{
		len+=x[i];
		len=max(len,0LL);
		if(len>=mid)
			len=0,++tot;
	}
	if(len>=mid)
		len=0,++tot;
	return tot<=k;
}
ll solvemin()
{
	ll L=1,R=1e18;
	ll ans=-1;
	while(L<=R)
	{
		ll mid=(L+R)>>1;
		if(checkmin(mid))
			ans=mid,R=mid-1;
		else
			L=mid+1;
	}
	return ans;
}
bool checkmax(ll mid)
{
	int tot=0;
	ll len=0;
	for(int i=1;i<=n;++i)
	{
		len+=x[i];
		len=max(len,0LL);
		if(len>=mid)
			len=0,++tot;
	}
	if(len>=mid)
		len=0,++tot;
	return tot>=k;
}
ll solvemax()
{
	ll L=1,R=1e18;
	ll ans=-1;
	while(L<=R)
	{
		ll mid=(L+R)>>1;
		if(checkmax(mid))
			ans=mid,L=mid+1;
		else
			R=mid-1;
	}
	return ans;
}
int main()
{
	n=read(),k=read();
	for(int i=1;i<=n;++i)
		x[i]=read();
	ll a=solvemin(),b=solvemax();
	if(a>b || a<0 || b<0)
		puts("-1");
	else
		cout<<a<<' '<<b<<endl;
	return 0;
}