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.
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 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).
3 2
1 2 1
1 2
1 3
3
6
8 3
1 1 2 2 1 3 1 1
2 7
1 6
2 7
20
20
20
Consider the following array (see the second sample) and its [2, 7] subarray (elements of the subarray are colored):
题意：
给你一个序列an 。t次询问，问在[L, R] 的区间里面的值是多少。公式是：ai的个数的平方 * ai 的求和。具体看Note。
题解：
这一题用莫队算法。如果做过小Z的袜子的话，这一题很简单，小Z的袜子是 ai个数的平方求和，这里就乘多了一个ai。
巨坑：num[] 数组要用int 开，用long long开的话会被卡TLE。（很绝望，被卡了10次罚时，最后用输入挂才过的）
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <string> 5 #include <algorithm> 6 #include <cmath> 7 #include <vector> 8 #include <queue> 9 #include <map> 10 #include <stack> 11 #include <set> 12 using namespace std; 13 typedef long long LL; 14 #define ms(a, b) memset(a, b, sizeof(a)) 15 #define pb push_back 16 #define mp make_pair 17 const LL INF = 0x7fffffff; 18 const int inf = 0x3f3f3f3f; 19 const int mod = 1e9+7; 20 const int maxn = 1000000+10; 21 22 template<class T> 23 inline bool scan_d(T &ret){ 24 char c;int sgn; 25 if(c=getchar(), c==EOF) return 0; 26 while(c!='-'&&(c<'0'||c>'9')) c = getchar(); 27 sgn=(c=='-')?-1:1; 28 ret=(c=='-')?0:(c-'0'); 29 while(c=getchar(), c>='0'&&c<='9') ret = ret*10+(c-'0'); 30 ret*=sgn; 31 return 1; 32 } 33 inline void out(LL x) 34 { 35 if(x>9) out(x/10); 36 putchar(x%10+'0'); 37 } 38 39 int a[maxn]; 40 int unit, n, t; 41 LL ans[maxn]; 42 int num[maxn]; 43 44 45 46 struct node 47 { 48 int l, r, id; 49 }que[maxn]; 50 void init() { 51 ms(num, 0); 52 } 53 bool cmp(node x1, node x2) 54 { 55 if(x1.l/unit != x2.l/unit){ 56 return x1.l/unit < x2.l/unit; 57 } 58 else{ 59 return x1.r < x2.r; 60 } 61 } 62 void work() 63 { 64 sort(que, que+t, cmp); 65 LL temp = 0; 66 int l, r; 67 l = 0, r = 0; 68 for(int i =0;i<t;i++){ 69 while(r<que[i].r){ 70 r++; 71 temp -= 1LL*num[a[r]]*num[a[r]]*a[r]; 72 num[a[r]]++; 73 temp += 1LL*num[a[r]]*num[a[r]]*a[r]; 74 } 75 while(r>que[i].r){ 76 temp -= 1LL*num[a[r]]*num[a[r]]*a[r]; 77 num[a[r]]--; 78 temp += 1LL*num[a[r]]*num[a[r]]*a[r]; 79 r--; 80 } 81 while(l>que[i].l){ 82 l--; 83 temp -= 1LL*num[a[l]]*num[a[l]]*a[l]; 84 num[a[l]]++; 85 temp += 1LL*num[a[l]]*num[a[l]]*a[l]; 86 } 87 while(l<que[i].l){ 88 temp -= 1LL*num[a[l]]*num[a[l]]*a[l]; 89 num[a[l]]--; 90 temp += 1LL*num[a[l]]*num[a[l]]*a[l]; 91 l++; 92 } 93 ans[que[i].id] = temp; 94 } 95 } 96 void solve() { 97 // scanf("%d%d", &n, &t); 98 scan_d(n);scan_d(t); 99 for(int i = 1;i<=n;i++) 100 scan_d(a[i]); 101 unit = (int)sqrt(n); 102 for(int i = 0;i<t;i++){ 103 que[i].id = i; 104 // scanf("%d%d", &que[i].l, &que[i].r); 105 scan_d(que[i].l); 106 scan_d(que[i].r); 107 } 108 work(); 109 for(int i = 0;i<t;i++){ 110 // cout << ans[i] << endl; 111 // printf("%I64d ", ans[i]); 112 out(ans[i]); 113 puts(""); 114 } 115 } 116 int main() { 117 #ifdef LOCAL 118 freopen("input.txt", "r", stdin); 119 // freopen("output.txt", "w", stdout); 120 #endif 121 // ios::sync_with_stdio(0); 122 // cin.tie(0); 123 init(); 124 solve(); 125 return 0; 126 }