cf86d

cf86d

​​ D. Powerful array time limit per test 5 seconds

memory limit per test 256 megabytes

input standard input

output standard output

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).

Examples 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 Note Consider the following array (see the second sample) and its [2, 7] subarray (elements of the subarray are colored):

Then K1 = 3, K2 = 2, K3 = 1, so the power is equal to 32·1 + 22·2 + 12·3 = 20.

这题最后输出要用i64d才行，lld会T掉。 题意求【l,r】中各个数出现次数的平方×这个数的数值，其实就是小b的询问多加了点东西，这题卡常。。

#include#include#include#define LL long long#define N 220000inline int read(){ int x=0;char ch=getchar(); while (ch<'0'||ch>'9') ch=getchar(); while (ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=getchar();} return x;}struct node{ int l,r,id,bl;}q[N];int n,t,n1;long long f[N<<2+1],a[N];inline bool cmp(node a,node b){ return a.bl==b.bl?a.rl) --l1,tmp+=((f[a[l1]]<<1)+1)*a[l1],f[a[l1]]++; while (rr1) ++r1,tmp+=((f[a[r1]]<<1)+1)*a[r1],f[a[r1]]++; ans[id]=tmp; } for (int i=1;i<=t;++i) printf("%I64d\n",ans[i]); return 0;}

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