CF86 D. Powerful array

题目传送门:https://codeforces.com/problemset/problem/86/D

题目大意:
给定一个长度为\(n\)的序列,有\(m\)组询问,每次询问\([l,r]\)中,\(\sum\limits_{s}K_s^2\times s\)的值,其中,\(K_s\)表示\(s\)在子串\([l,r]\)中的出现次数


由于这题没有修改,仅有询问,故我们可以考虑莫队算法

/*program from Wolfycz*/
#include<map>
#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define Fi first
#define Se second
#define ll_inf 1e18
#define MK make_pair
#define sqr(x) ((x)*(x))
#define pii pair<int,int>
#define int_inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
	static char buf[1000000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
template<typename T>inline T frd(T x){
	int f=1; char ch=gc();
	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')    f=-1;
	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
template<typename T>inline T read(T x){
	int f=1; char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x<0)	putchar('-'),x=-x;
	if (x>9)	print(x/10);
	putchar(x%10+'0');
}
const int N=2e5,M=1e6;
int A[N+10],pos[N+10];
struct node{
	int l,r,ID;
	void Read(int i){l=read(0),r=read(0),ID=i;}
	node(int _l=0,int _r=0,int _ID=0){l=_l,r=_r,ID=_ID;}
	bool operator <(const node &tis)const{return pos[l]!=pos[tis.l]?l<tis.l:r<tis.r;}
}B[N+10];
ll All,Ans[N+10],Cnt[M+10];
void Add(int x,int v){
	All-=1ll*x*sqr(Cnt[x]);
	Cnt[x]+=v;
	All+=1ll*x*sqr(Cnt[x]);
}
int main(){
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	int n=read(0),m=read(0),size=sqrt(n);
	for (int i=1;i<=n;i++)	A[i]=read(0),pos[i]=(i-1)/size+1;
	for (int i=1;i<=m;i++)	B[i].Read(i);
	sort(B+1,B+1+m);
	for (int i=1,l=1,r=0;i<=m;i++){
		while (r<B[i].r)	Add(A[++r], 1);
		while (r>B[i].r)	Add(A[r--],-1);
		while (l<B[i].l)	Add(A[l++],-1);
		while (l>B[i].l)	Add(A[--l], 1);
		Ans[B[i].ID]=All;
//		printf("\n");
	}
	for (int i=1;i<=m;i++)	printf("%lld\n",Ans[i]);
	return 0;
}
posted @ 2021-07-01 21:40  Wolfycz  阅读(18)  评论(0编辑  收藏  举报