Codeforces#86D Powerful array（分块暴力）

最新推荐文章于 2023-03-29 23:41:56 发布

weixin_34123613

最新推荐文章于 2023-03-29 23:41:56 发布

阅读量67

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/clnchanpin/p/6732957.html

本文介绍了一个算法问题，即计算给定数组中特定子数组的能量值，能量定义为子数组中各元素出现频率的平方乘以其值的总和。文章提供了一种通过分块暴力的方法来高效解决该问题，并附带了完整的C++实现代码。

Description

An array of positive integers a₁, a₂, ..., a_n is given. Let us consider its arbitrary subarray a_l, a_l + 1..., a_r, where 1 ≤ l ≤ r ≤ n. For every positive integer s denote by K_s the number of occurrences of s into the subarray. We call the power of the subarray the sum of products K_s·K_s·s for every positive integer s. The sum contains only finite number of nonzero summands as the number of different values in the array is indeed finite.

You should calculate the power of t given subarrays.

Input

First line contains two integers n and t (1 ≤ n, t ≤ 200000) — the array length and the number of queries correspondingly.

Second line contains n positive integers a_i (1 ≤ a_i ≤ 10⁶) — the elements of the array.

Next t lines contain two positive integers l, r (1 ≤ l ≤ r ≤ n) each — the indices of the left and the right ends of the corresponding subarray.

Output

Output t lines, the i-th line of the output should contain single positive integer — the power of the i-th query subarray.

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preferred to use cout stream (also you may use %I64d).

Sample Input

Input

Output

3
6

Input

8 3
1 1 2 2 1 3 1 1
2 7
1 6
2 7

Output

20
20
20

题意：就问问你统计[L,R]中x1,x2,,xi出现次数a1,a2,,ai的ai*ai*xi和。

题解：分块暴力。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<set>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef __int64 LL;
const int INF = 0x3f3f3f3f;
const int maxn=200000+100;
LL up[maxn];
int col[maxn],sum[maxn*10],n,m,k;
LL ans;
struct node{
    int l,r;
    int id;
}q[maxn];
int cmp(node l1,node l2)
{
    if(l1.l/k==l2.l/k)
        return l1.r<l2.r;
    return l1.l<l2.l;
}
void update(int x,int val)
{
    ans-=(LL)col[x]*sum[col[x]]*sum[col[x]];
    sum[col[x]]+=val;
    ans+=(LL)col[x]*sum[col[x]]*sum[col[x]];
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        CLEAR(sum,0);
        k=sqrt(n*1.0+0.5);
        REPF(i,1,n)
            scanf("%d",&col[i]);
        REP(i,m)
        {
            scanf("%d%d",&q[i].l,&q[i].r);
            q[i].id=i;
        }
        sort(q,q+m,cmp);
        int L=0,R=0;
        ans=0;
        for(int i=0;i<m;i++)
        {
            int id=q[i].id;
            int l=q[i].l;
            int r=q[i].r;
            while(L < l)
            {
                update(L,-1);
                L++;
            }
            while(R > r)
            {
                update(R,-1);
                R--;
            }
            while(L > l)
            {
                L--;
                update(L,1);
            }
            while(R < r)
            {
                R++;
                update(R,1);
            }
            up[id]=ans;
        }
        for(int i=0;i<m;i++)
            printf("%I64d\n",up[i]);
    }
    return 0;
}

转载于:https://www.cnblogs.com/clnchanpin/p/6732957.html