Codeforces 86D

题目大意：

给定长为N的序列a[1..N]和T次询问，每次询问指定一个子段a[L..R]，设某个数v在该子段中出现k(v)次，那么该次询问的结果为Σv*k(v)*k(v)。

N,T≤2e5，1≤a[i]≤1e6。

莫队算法初次听起来可能感觉很神奇，不过原理和实现方法都很简单。这里有篇很不错的英文讲解：https://www.hackerearth.com/zh/practice/notes/mos-algorithm/

应用莫队算法的前提：（1）静态序列，无修改操作；（2）离线询问。

莫队算法的流程：（直接从上边的网址Ctrl-C过来的）

Step 1:
    Denote BLOCK_SIZE = sqrt(N);
Step 2:
    Rearrange all queries in a way we will call “Mo’s order”. It is defined like this: [L1, R1] comes earlier than [L2, R2] in Mo’s order if and only if:
   a) L1 / BLOCK_SIZE < L2 / BLOCK_SIZE
   b) L1 / BLOCK_SIZE == L2 / BLOCK_SIZE && R1 < R2
Step 3:
    Maintain segment [mo_left, mo_right] for which we know Func([mo_left, mo_right]). Initially, this segment is empty. We set mo_left = 0 and mo_right = -1;（注：原文约定序列下标从0开始）
Step 4:
    Answer all queries following Mo’s order. Suppose the next query you want to answer is [L, R]. Then you perform these steps:
   a) while mo_right is less than R, extend current segment to [mo_left, mo_right + 1];
   b) while mo_right is greater than R, cut current segment to [mo_left, mo_right - 1];
   c) while mo_left is greater than L, extend current segment to [mo_left - 1, mo_right];
   d) while mo_left is less than L, cut current segment to [mo_left + 1, mo_right].

本题的代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>

using LL = long long;

const int blockSize = 450;
const int maxN = (int)2e5 + 10;

int A[maxN], N, T;
int C[(int)1e6 + 1];
LL curAns;

struct Query
{
    int id;
    int left, right;
    int blockId;
    LL ans;

    bool operator < (const Query& rhs) const {
        return blockId < rhs.blockId || (blockId == rhs.blockId && right < rhs.right);
    }
};
Query query[maxN];

void input()
{
    scanf("%d%d", &N, &T);
    for (int i = 1; i <= N; i++)
        scanf("%d", A + i);
    for (int l, r, i = 1; i <= T; i++)
    {
        scanf("%d%d", &l, &r);
        query[i] = {i, l, r, l / blockSize, 0LL};
    }
    std::sort(query + 1, query + T + 1);
}

inline void add(int pos)
{
    int &c = C[A[pos]];
    curAns += (2LL * c + 1) * A[pos];
    c += 1;
}
inline void subtract(int pos)
{
    int &c = C[A[pos]];
    curAns -= (2LL * c - 1) * A[pos];
    c -= 1;
}

void process()
{
    int left = 0, right = 0;
    for (int i = 1; i <= T; i++)
    {
        while (left < query[i].left)
            subtract(left++);
        while (left > query[i].left)
            add(--left);
        while (right < query[i].right)
            add(++right);
        while (right > query[i].right)
            subtract(right--);

        query[i].ans = curAns;
    }
}

void solve()
{
    process();
    std::sort(query + 1, query + T + 1, [] (const Query& A, const Query& B) {
        return A.id < B.id;
    });
    for (int i = 1; i <= T; i++)
        printf("%lld
", query[i].ans);
}

int main()
{
    input();
    solve();
    return 0;
}