[HRBUSTOJ1476]Pairs（FFT）

题目链接：http://acm-software.hrbust.edu.cn/problem.php?id=1476

题意：给n个数，m次询问，每次询问一个k。问n个数里两数之和严格小于k的数对。

根据输入样例，无非是需要求：

f = cnt(1 2 3 4 5)T * (x)（其中(x)代表1,x,x^2...的向量，cnt表示这些数出现的次数，T表示转置）自乘一次后对应x幂次前的系数，取前k-1项系数和就是答案。复杂度太高了，O(n^2)的系数搞法是不可以的，所以需要搞成点值表达DFT后再IDFT回来。

#include <bits/stdc++.h>
using namespace std;

const double PI = acos(-1.0);
typedef struct Complex {
    double r,i;
    Complex(double _r = 0.0,double _i = 0.0) {
        r = _r; i = _i;
    }
    Complex operator +(const Complex &b) {
        return Complex(r+b.r,i+b.i);
    }
    Complex operator -(const Complex &b) {
        return Complex(r-b.r,i-b.i);
    }
    Complex operator *(const Complex &b) {
        return Complex(r*b.r-i*b.i,r*b.i+i*b.r);
    }
}Complex;

void change(Complex y[],int len) {
    int i,j,k;
    for(i = 1, j = len/2;i < len-1; i++) {
        if(i < j)swap(y[i],y[j]);
        k = len/2;
        while( j >= k) {
            j -= k;
            k /= 2;
        }
        if(j < k) j += k;
    }
}

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].r /= len;
        }
    }
}

const int maxn = 400040;
typedef long long LL;
int a[maxn];
Complex c[maxn];
LL f[maxn];
int n, m;

int main() {
    // freopen("in", "r", stdin);
    int T;
    scanf("%d", &T);
    while(T--) {
        memset(f, 0, sizeof(f));
        scanf("%d%d",&n,&m);
        int maxx = -1;
        for(int i = 0; i < n; i++) {
            scanf("%d", &a[i]);
            maxx = max(a[i], maxx);
            f[a[i]]++;
        }
        int len1 = maxx + 1;
        int len = 1;
        while(len < 2 * len1) len <<= 1;
        for(int i = 0; i < len; i++) c[i] = Complex(0, 0);
        for(int i = 0; i < len1; i++) c[i] = Complex(f[i], 0);
        fft(c, len, 1);
        for(int i = 0; i < len; i++) c[i] = c[i] * c[i];
        fft(c, len, -1);
        for(int i = 0; i < len; i++) f[i] = (LL)(c[i].r + 0.5);
        len = 2 * maxx;
        for(int i = 0; i < n; i++) f[a[i]*2]--;
        for(int i = 1; i <= len; i++) {
            f[i] /= 2;
            f[i] += f[i-1];
        }
        while(m--) {
            int k;
            scanf("%d", &k);
            printf("%lld\n", f[k-1]);
        }
    }
    return 0;
}