CC Arithmetic Progressions (FFT + 分块处理)

学习的cxlove大神的博客：传送门

关键点：枚举中间的数，如果其他两个数不在当前块中，那么前面的所有的块和后面所有的块做卷积，得到前面序列和后面序列相乘的情况。如果有这两个数有数在当前块，那么枚举前一个数在当前块，后一个数不在当前块，枚举后一个数在当前块，前一个数在当前块或者不在当前块，都可以。

分块的关键：是复杂度降到o(k∗(Nk∗Nk+M∗logM)这里不能按照sqrt(N)来分块，复杂度过大。

代码：

#include <bits/stdc++.h>#define LL long long#define FOR(i,x,y)  for(int i = x;i < y;++ i)#define IFOR(i,x,y) for(int i = x;i > y;-- i)using namespace std;//FFT copy from kuangbinconst double pi = acos (-1.0);// Complex  z = a + b * istruct Complex {    double a, b;    Complex(double _a=0.0,double _b=0.0):a(_a),b(_b){}    Complex operator + (const Complex &rhs) const {        return Complex(a + rhs.a , b + rhs.b);    }    Complex operator - (const Complex &rhs) const {        return Complex(a - rhs.a , b - rhs.b);    }    Complex operator * (const Complex &rhs) const {        return Complex(a * rhs.a - b * rhs.b , a * rhs.b + b * rhs.a);    }};//len = 2 ^ kvoid 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 IDFTvoid 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 maxn = 100010;const int maxm = 30030;int l[maxm<<2],r[maxm<<2];LL num[maxm<<2];int a[maxn],n,len;void init(){    memset(l,0,sizeof(l));    memset(r,0,sizeof(r));    int mx_len = -1;    FOR(i,0,n){        scanf("%d",&a[i]);        ++ r[a[i]];        mx_len = max(mx_len,a[i]);    }    mx_len = (mx_len+1) << 1;    len = 1;    while(len < mx_len) len <<= 1;}Complex x1[maxm<<2],x2[maxm<<2],x[maxm<<2];void calc(){    FOR(i,0,len)    x1[i] = Complex(l[i],0);//偷懒写法，这个时候l[maxm<<2]，这里一直re，其实不应该偷懒的    FOR(i,0,len)    x2[i] = Complex(r[i],0);    FFT(x1,len,1);    FFT(x2,len,1);    FOR(i,0,len)    x[i] = x1[i]*x2[i];    FFT(x,len,-1);    FOR(i,0,len)    num[i] = (LL)(x[i].a+0.5);}void work(){    LL ans = 0;    int blocks = min(n,30);    int blk = (n+blocks-1)/blocks;    FOR(i,0,blocks){        int s = i*blk,t = min(n,(i+1)*blk);        FOR(j,s,t)  -- r[a[j]];        calc();        FOR(j,s,t){            int y = (a[j]<<1);            ans += num[y];            FOR(k,s,j) if(y >= a[k] && y-a[k] < maxm)                ans += r[y-a[k]];            FOR(k,j+1,t) if(y >= a[k] && y-a[k] < maxm)                ans += l[y-a[k]];            ++ l[a[j]];        }    }    printf("%lld\n",ans);}int main(){    //freopen("test.in","r",stdin);    while(~scanf("%d",&n)){        init();        work();    }}