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], belong[MAXN], mxblock, num[MAXN];
LL Solve1(int l, int r) {
for(int i = 1; i < l; i++) num[a[i]]++;
LL ret = 0;
for(int j = l; j <= r; j++)
for(int k = j + 1; k <= r; k++)
if(2 * a[j] - a[k] >= 0) ret += num[2 * a[j] - a[k]];
for(int i = 1; i < l; i++) num[a[i]]--;
for(int i = N; i > r; i--) num[a[i]]++;
for(int j = r; j >= l; j--)
for(int k = j - 1; k >= l; k--)
if(2 * a[j] - a[k] >= 0) ret += num[2 * a[j] - a[k]];
for(int i = N; i > r; i--) num[a[i]]--;
for(int j = l; j <= r; j++) {
for(int i = j - 1; i >= l; i--) num[a[i]]++;
for(int k = j + 1; k <= r; k++)
if(2 * a[j] - a[k] >= 0) ret += num[2 * a[j] - a[k]];
for(int i = j - 1; i >= l; i--) num[a[i]]--;
}
return ret;
}
int ta[MAXN], tb[MAXN], Lim;
LL Solve2(int l, int r) {
memset(ta, 0, sizeof(ta));
memset(tb, 0, sizeof(tb));
for(int i = l - 1; i >= 1; i--) ta[a[i]]++;
for(int i = r + 1; i <= N; i++) tb[a[i]]++;
Mul(ta, tb, mx, mx); LL ret = 0;
for(int i = l; i <= r; i++) ret += tb[2 * a[i]];
return ret;
}
signed main() {
//freopen("a.in", "r", stdin);
N = read(); block = sqrt(8 * N * log2(N));
memset(ll, 0x3f, sizeof(ll));
for(int i = 1; i <= N; i++) {
belong[i] = (i - 1) / block + 1; chmax(mxblock, belong[i]);
chmin(ll[belong[i]], i);
chmax(rr[belong[i]], i);
a[i] = read(), chmax(mx, a[i]);
}
Lim = GetLen(mx); Init(4 * Lim);
LL ans = 0;
for(int i = 1; i <= mxblock; i++) {
ans += Solve1(ll[i], rr[i]);
ans += Solve2(ll[i], rr[i]);
}
cout << ans;
return 0;
}
/*
7
7 0 4 7 0 8 8
*/