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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> Codeforces#86D Powerful array(分塊暴力)

Codeforces#86D Powerful array(分塊暴力)

編輯:C++入門知識

Codeforces#86D Powerful array(分塊暴力)


Description

An array of positive integers a1, a2, ..., an is given. Let us consider its arbitrary subarray al, al + 1..., ar, where 1 ≤ l ≤ r ≤ n. For every positive integer s denote by Ks the number of occurrences of s into the subarray. We call the power of the subarray the sum of products Ks·Ks·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 ai (1 ≤ ai ≤ 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
3 2
1 2 1
1 2
1 3
Output
3
6
Input
8 3
1 1 2 2 1 3 1 1
2 7
1 6
2 7
Output
20
20
20

題意:就問問你統計[L,R]中x1,x2,,xi出現次數a1,a2,,ai的ai*ai*xi和。

題解:分塊暴力。

 

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
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 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

 

 

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