CodeForces-86D Powerful array

題意: 給定一個區間計算power值, power[l, r] = sigma(sum[pi]*sum[pi]*pi).
分析: 莫隊來做, 對於add操作, 要累加的值為(sum[a[x]]*2 + 1)*a[x], del操作減去就行. 這個規律很好發現. 這道題卡常, 乘法用位操作替代. 輸出一定要I64d, 否則TLE.
程式碼:

#include <bits/stdc++.h>

using namespace std;

const int MAXN = 2e5 + 10;
const int MAXM = 1e6 + 10;
struct Query
{
    int l, r, id;
} Q[MAXN];

int L = 1, R = 0;
long long Ans = 0;
long long ans[MAXN];
int flag[MAXM], pos[MAXN], a[MAXN];
int n, m;
bool cmp(Query x, Query y)
{
    if (pos[x.l] == pos[y.l])
    {
        return x.r < y.r;
    }
    else
    {
        return pos[x.l] < pos[y.l];
    }
}
inline 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;
}

inline void add(int x)
{

    Ans += (((long long)(flag[a[x]]) << 1) + 1) * a[x];
    flag[a[x]]++;
    return;
}

inline void del(int x)
{
    flag[a[x]]--;
    Ans -= (((long long)(flag[a[x]]) << 1) + 1) * a[x];
    return;
}
int main()
{
    // cin.sync_with_stdio(false);
    // cin.tie(0);
    //scanf("%d%d", &n, &m);
    //cin >> n >> m;
    n = read();
    m = read();
    int unit = sqrt(1.0 * n);
    for (int i = 1; i <= n; i++)
    {
        //cin >> a[i];
        a[i] = read();
        pos[i] = i / unit + 1;
    }

    for (int i = 1; i <= m; i++)
    {
        Q[i].l = read();
        Q[i].r = read();
        Q[i].id = i;
    }

    sort(Q + 1, Q + m + 1, cmp);

    for (int i = 1; i <= m; i++)
    {
        if (Q[i].l == Q[i].r)
        {
            ans[Q[i].id] = 1l * a[Q[i].l];
            continue;
        }
        while (L < Q[i].l)
        {
            del(L++);
            //L++;
        }

        while (L > Q[i].l)
        {
            //L--;
            add(--L);
        }

        while (R < Q[i].r)
        {
            //R++;
            add(++R);
        }

        while (R > Q[i].r)
        {
            del(R--);
            //R--;
        }
        ans[Q[i].id] = Ans;
    }

    for (int i = 1; i <= m; i++)
    {
        printf("%I64d\n", ans[i]);
        // cout << ans[i] << endl;
    }
    return 0;
}