CC Arithmetic Progressions (FFT + 分块处理)
转载请注明出处，谢谢http://blog.csdn.net/ACM_cxlove?viewmode=contents by---cxlove
题目：给出n个数，选出三个数，按下标顺序形成等差数列
http://www.codechef.com/problems/COUNTARI
如果只是形成 等差数列并不难，大概就是先求一次卷积，然后再O(n)枚举，判断2 * a[i]的种数，不过按照下标就不会了。
有种很矬的，大概就是O(n)枚举中间的数，然后 对两边分别卷积，O(n * n * lgn)。
如果能想到枚举中间的数的话，应该可以进一步想到分块处理。
如果分为K块
那么分为几种情况 ：
三个数都是在当前块中，那么可以枚举后两个数，查找第一个数，复杂度O(N/K * N/K)
两个数在当前块中，那么另外一个数可能在前面，也可能在后面，同理还是枚举两个数，查找，复杂度
O(N/K * N/K)
如果只有一个数在当前块中，那么就要对两边的数进行卷积，然后枚举当前块中的数，查询2 × a[i]。复杂度O(N * lg N)
那么总体就是O(k * (N/K * N/K + N * lg N))。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
//FFT copy from kuangbin
const double pi = acos (-1.0);
// Complex z = a + b * i
struct Complex {
double a, b;
Complex(double _a=0.0,double _b=0.0):a(_a),b(_b){}
Complex operator + (const Complex &c) const {
return Complex(a + c.a , b + c.b);
}
Complex operator - (const Complex &c) const {
return Complex(a - c.a , b - c.b);
}
Complex operator * (const Complex &c) const {
return Complex(a * c.a - b * c.b , a * c.b + b * c.a);
}
};
//len = 2 ^ k
void change (Complex y[] , int len) {
for (int i = 1 , j = len / 2 ; i < len -1 ; i ++) {
if (i < j) swap(y[i] , y[j]);
int k = len / 2;
while (j >= k) {
j -= k;
k /= 2;
}
if(j < k) j += k;
}
}
// FFT
// len = 2 ^ k
// on = 1 DFT on = -1 IDFT
void FFT (Complex y[], int len , int on) {
change (y , len);
for (int h = 2 ; h <= len ; h <<= 1) {
Complex wn(cos (-on * 2 * pi / h), sin (-on * 2 * pi / h));
for (int j = 0 ; j < len ; j += h) {
Complex w(1 , 0);
for (int k = j ; k < j + h / 2 ; k ++) {
Complex u = y[k];
Complex t = w * y [k + h / 2];
y[k] = u + t;
y[k + h / 2] = u - t;
w = w * wn;
}
}
}
if (on == -1) {
for (int i = 0 ; i < len ; i ++) {
y[i].a /= len;
}
}
}
const int N = 100005;
typedef long long LL;
int n , a[N];
int block , size;
LL num[N << 2];
int min_num = 30000 , max_num = 1;
int before[N] = {0}, behind[N] = {0} , in[N] = {0};
Complex x1[N << 2] ,x2[N << 2];
int main () {
#ifndef ONLINE_JUDGE
freopen("input.txt" , "r" , stdin);
#endif
scanf ("%d", &n);
for (int i = 0 ; i < n ; ++ i) {
scanf ("%d", &a[i]);
behind[a[i]] ++;
min_num = min (min_num , a[i]);
max_num = max (max_num , a[i]);
}
LL ret = 0;
block = min(n , 35);
size = (n + block - 1) / block;
for (int t = 0 ; t < block ; t ++) {
int s = t * size , e = (t + 1) * size;
if (e > n) e = n;
for (int i = s ; i < e ; i ++) {
behind[a[i]] --;
}
for (int i = s ; i < e ; i ++) {
for (int j = i + 1 ; j < e ; j ++) {
int m = 2 * a[i] - a[j];
if(m >= 1 && m <= 30000) {
// both of three in the block
ret += in[m];
// one of the number in the pre block
ret += before[m];
}
m = 2 * a[j] - a[i];
if (m >= 1 && m <= 30000) {
// one of the number in the next block
ret += behind[m];
}
}
in[a[i]] ++;
}
// pre block , current block , next block
if (t > 0 && t < block - 1) {
int l = 1;
int len = max_num + 1;
while (l < len * 2) l <<= 1;
for (int i = 0 ; i < len ; i ++) {
x1[i] = Complex (before[i] , 0);
}
for (int i = len ; i < l ; i ++) {
x1[i] = Complex (0 , 0);
}
for (int i = 0 ; i < len ; i ++) {
x2[i] = Complex (behind[i] , 0);
}
for (int i = len ; i < l ; i ++) {
x2[i] = Complex (0 , 0);
}
FFT (x1 , l , 1);
FFT (x2 , l , 1);
for (int i = 0 ; i < l ; i ++) {
x1[i] = x1[i] * x2[i];
}
FFT (x1 , l , -1);
for (int i = 0 ; i < l ; i ++) {
num[i] = (LL)(x1[i].a + 0.5);
}
for (int i = s ; i < e ; i ++) {
ret += num[a[i] << 1];
}
}
for (int i = s ; i < e ; i ++) {
in[a[i]] --;
before[a[i]] ++;
}
}
printf("%lld\n", ret);
return 0;
}