【JZOJ6434】【luoguP5665】【CSP-S2019】划分

description

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analysis

• 首先有一个结论，对于\([1,i]\)区间划分最后一段的和尽量小，答案会更优，具体证明参考毛爷爷的博客

• 设\(f[i]\)为满足\([1,i]\)划分最优时、\((f[i],i]\)这段和最小时的最右的端点，最优划分即为从\(n\)开始向\(f\)不断统计

• 由后一段比前一段大可知\(sum[f[i]]-sum[f[f[i]]]≤sum[i]-sum[f[i]]\)，即\(sum[i]≥2sum[f[i]]-sum[f[f[i]]]\)

• 右边只和\(f[i]\)有关，把右边的记为\(clac(f[i])\)，于是可以维护一个\(clac\)值上升的单调队列来求出每一位的\(f\)

• 具体维护就是，队头不断出队直到\(sum[i]<clac(q[head])\)（因为\(f[i]\)要尽可能大），\(f[i]\)赋为队头

• 若\(clac(i)≤clac(q[tail])\)队尾再不断出队（因为\(clac(i)\)更小且\(i\)更靠右），最后\(i\)进队，就完成了队列的维护

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code

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 40000005
#define ha 1073741824
#define ll long long
#define LL __int128
#define reg register int
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)

using namespace std;

int p[100005],l[100005],r[100005];
int f[MAXN],q[MAXN];
ll b[MAXN],sum[MAXN];
int n,type;

inline int read()
{
	ll x=0,f=1;char ch=getchar();
	while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
	while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
inline ll clac(ll x){return 2*sum[x]-sum[f[x]];}
inline void print(LL x)
{
	int num[55];num[0]=0;
	while (x)num[++num[0]]=x%10,x/=10;
	while (num[0])printf("%d",num[num[0]--]);
	printf("\n");
}
int main()
{
	n=read(),type=read();
	if (type)
	{
		int x=read(),y=read(),z=read();b[1]=read(),b[2]=read();int m=read(),now=1;
		fo(i,1,m)p[i]=read(),l[i]=read(),r[i]=read();
		fo(i,3,n)b[i]=(x*b[i-1]%ha+y*b[i-2]%ha+z)%ha;
		fo(i,1,n)
		{
			if (i>p[now])++now;
			sum[i]=sum[i-1]+(b[i]%(r[now]-l[now]+1))+l[now];
		}
	}
	else {fo(i,1,n)sum[i]=sum[i-1]+read();}
	int head=0,tail=0;
	fo(i,1,n)
	{
		while (head<tail && sum[i]>=clac(q[head+1]))++head;
		f[i]=q[head];
		while (head<tail && clac(i)<=clac(q[tail]))--tail;
		q[++tail]=i;
	}
	LL ans=0;
	for (reg i=n;i;i=f[i])ans+=(LL)(sum[i]-sum[f[i]])*(sum[i]-sum[f[i]]);
	print(ans);
	return 0;
}