题目链接

BZOJ3509

题解

化一下式子，就是
\[2A[j] = A[i] + A[k]\]
所以我们对一个位置两边的数构成的生成函数相乘即可
但是由于这样做是\(O(n^2logn)\)的，我们考虑如何优化

显然可以分块做，我们不对所有数左右求卷积，只对\(B\)个块左右做，这样\(i\)和\(k\)都在块外的情况就可以统计出来
\(i\)或\(k\)在块内的情况可以暴力扫一遍计算
复杂度\(O(Bnlogn + nB)\)
经计算\(B = \sqrt{nlogn}\)最优
但考虑到\(fft\)的常数问题，\(B = 2000\)左右比较合理
复杂度就大概是\(O((nlogn)^{\frac{3}{2}})\)
竟然能\(A\)....
交上去就垫底了。。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define REP(i,n) for (register int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cp pair<int,int>
#define LL long long int
#define res register
using namespace std;
const int maxn = 400005,maxm = 4005,INF = 1000000000;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
struct E{
double a,b;
E(){}
E(double x,double y):a(x),b(y) {}
E(int x,int y):a(x),b(y) {}
inline E operator =(const int& b){
this->a = b; this->b = 0;
return *this;
}
inline E operator =(const double& b){
this->a = b; this->b = 0;
return *this;
}
inline E operator /=(const double& b){
this->a /= b; this->b /= b;
return *this;
}
};
inline E operator *(const E& a,const E& b){
return E(a.a * b.a - a.b * b.b,a.a * b.b + a.b * b.a);
}
inline E operator *=(E& a,const E& b){
return a = E(a.a * b.a - a.b * b.b,a.a * b.b + a.b * b.a);
}
inline E operator +(const E& a,const E& b){
return E(a.a + b.a,a.b + b.b);
}
inline E operator -(const E& a,const E& b){
return E(a.a - b.a,a.b - b.b);
}
const double pi = acos(-1);
int R[maxn];
void fft(E* a,int n,int f){
for (res int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]);
for (res int i = 1; i < n; i <<= 1){
E wn(cos(pi / i),f * sin(pi / i));
for (res int j = 0; j < n; j += (i << 1)){
E w(1,0),x,y;
for (res int k = 0; k < i; k++,w = w * wn){
x = a[j + k],y = w * a[j + k + i];
a[j + k] = x + y; a[j + k + i] = x - y;
}
}
}
if (f == -1) for (int i = 0; i < n; i++) a[i] /= n;
}
E A[maxn],C[maxn];
int N,B,val[maxn],b[maxn],bl[maxm],br[maxm],bi;
LL ans;
void work1(){
for (res int x = 2; x < bi; x++){
int deg1 = 0,deg2 = 0;
for (res int i = 1; b[i] != x; i++)
A[val[i]].a++,deg1 = max(deg1,val[i]);
for (res int i = N; b[i] != x; i--)
C[val[i]].a++,deg2 = max(deg2,val[i]);
int n = 1,L = 0;
while (n <= deg1 + deg2) n <<= 1,L++;
for (res int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
fft(A,n,1); fft(C,n,1);
for (res int i = 0; i < n; i++) A[i] *= C[i];
fft(A,n,-1);
for (res int i = bl[x]; i <= br[x]; i++)
ans += (LL)floor(A[val[i] << 1].a + 0.1);
for (res int i = 0; i < n; i++) A[i] = C[i] = 0.0;
}
}
int bac[maxn],M;
void work2(){
for (res int i = 1; i <= N; i++){
for (res int j = i + 1; b[j] == b[i]; j++){
if (val[j] >= val[i] && val[i] >= val[j] - val[i])
ans += bac[val[i] - (val[j] - val[i])];
if (val[j] < val[i] && val[i] + val[i] - val[j] <= 30000)
ans += bac[val[i] + val[i] - val[j]];
}
bac[val[i]]++;
}
REP(i,N) bac[val[i]]--;
for (res int i = N; i; i--){
for (res int j = i - 1; b[j] == b[i]; j--){
if (val[j] >= val[i] && val[i] >= val[j] - val[i])
ans += bac[val[i] - (val[j] - val[i])];
if (val[j] < val[i] && val[i] + val[i] - val[j] <= 30000)
ans += bac[val[i] + val[i] - val[j]];
}
bac[val[i]]++;
}
REP(i,N) bac[val[i]]--;
for (res int x = 1; x <= bi; x++){
for (res int i = bl[x]; i <= br[x]; i++){
for (res int j = i + 1; j <= br[x]; j++){
if (val[j] >= val[i] && val[i] >= val[j] - val[i])
ans -= bac[val[i] - (val[j] - val[i])];
if (val[j] < val[i] && val[i] + val[i] - val[j] <= 30000)
ans -= bac[val[i] + val[i] - val[j]];
}
bac[val[i]]++;
}
for (int i = bl[x]; i <= br[x]; i++) bac[val[i]]--;
}
}
int main(){
N = read(); B = 2000;
REP(i,N){
val[i] = read(),b[i] = i / B + 1,M = max(M,val[i]);
if (b[i] != b[i - 1]) br[b[i - 1]] = i - 1,bl[b[i]] = i;
}
br[b
] = N; bi = b
;
work1();
work2();
printf("%lld\n",ans);
return 0;
}