题意

链接

Sol

这都能分块。。。。

首先移一下项，变为统计多少\(i < j < k\)，满足\(2a[j] = a[i] + a[k]\)

发现\(a[i] \leqslant 30000\)，那么有一种暴力思路是枚举\(j\)，对于之前出现过的数构造一个生成函数，对于之后出现过的数构造一个生成函数，求一下第\(2a[j]\)项的值。复杂度\(O(nVlogV)\)

每次枚举\(j\)暴力卷积显然太zz了，我们可以分一下块，对于每一块之前之后的数分别构造生成函数暴力卷积算，对于块内的直接暴力(这里的暴力不只是统计块内的\((i, j, k)\)，还要考虑\(j, k\)在块内\(i\)在块外，以及\(i, j\)在块内，\(k\)在块外的情况，但都是可以暴力的)

如果分成\(B\)块的话，复杂度是\(\frac{N}{B} VlogV + B^2\)，假设\(n\)与\(V\)同阶的话，\(B\)大概取\(nlogn\)是最优的。此时复杂度为\(O(N \sqrt{Nlogn})\)

下面的代码在原BZOJ上可能会T

#include<bits/stdc++.h> 
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define LL long long 
#define ull unsigned long long 
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 1e6 + 10, INF = 1e9 + 1;
const double eps = 1e-9, pi = acos(-1);
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
namespace Poly {
    int rev[MAXN], GPow[MAXN], A[MAXN], B[MAXN], C[MAXN], lim, INV2;
    const int G = 3, mod = 1004535809, mod2 = 1004535808;
    template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
    template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
    template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
    template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
    template <typename A, typename B> inline bool chmax(A &x, B y) {return x < y ? x = y, 1 : 0;}
    template <typename A, typename B> inline bool chmin(A &x, B y) {return x > y ? x = y, 1 : 0;}
    int fp(int a, int p, int P = mod) {
        int base = 1;
        for(; p > 0; p >>= 1, a = 1ll * a * a % P) if(p & 1) base = 1ll * base *  a % P;
        return base;
    }
    int inv(int x) {
        return fp(x, mod - 2);
    }
    int GetLen(int x) {
        int lim = 1;
        while(lim < x) lim <<= 1;
        return lim;
    }
    void Init(/*int P,*/ int Lim) {
        INV2 = fp(2, mod - 2);
        for(int i = 1; i <= Lim; i++) GPow[i] = fp(G, (mod - 1) / i);
    }
    void NTT(int *A, int lim, int opt) {
        int len = 0; for(int N = 1; N < lim; N <<= 1) ++len; 
        for(int i = 1; i <= lim; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (len - 1));
        for(int i = 0; i <= lim; i++) if(i < rev[i]) swap(A[i], A[rev[i]]);
        for(int mid = 1; mid < lim; mid <<= 1) {
            int Wn = GPow[mid << 1];
            for(int i = 0; i < lim; i += (mid << 1)) {
                for(int j = 0, w = 1; j < mid; j++, w = mul(w, Wn)) {
                    int x = A[i + j], y = mul(w, A[i + j + mid]);
                    A[i + j] = add(x, y), A[i + j + mid] = add(x, -y);
                }
            }
        }
        if(opt == -1) {
            reverse(A + 1, A + lim);
            int Inv = fp(lim, mod - 2);
            for(int i = 0; i <= lim; i++) mul2(A[i], Inv);
        }
    }
    void Mul(int *a, int *b, int N, int M) {
        memset(A, 0, sizeof(A)); memset(B, 0, sizeof(B));
        int lim = 1, len = 0; 
        while(lim <= N + M) len++, lim <<= 1;
        for(int i = 0; i <= N; i++) A[i] = a[i]; 
        for(int i = 0; i <= M; i++) B[i] = b[i];
        NTT(A, lim, 1); NTT(B, lim, 1);
        for(int i = 0; i <= lim; i++) B[i] = mul(B[i], A[i]);
        NTT(B, lim, -1);
        for(int i = 0; i <= N + M; i++) b[i] = B[i];
        memset(A, 0, sizeof(A)); memset(B, 0, sizeof(B));
    }
};
using namespace Poly; 
int N, a[MAXN], mx, block, ll[MAXN], rr[MAXN], bel