CC Arithmetic Progressions (FFT + 分块处理)

分类： ACM_数学类 2013-07-25 21:27  505人阅读  评论(0)  收藏  举报

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;
 }