[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 = , j = len/;i < len-; i++) {
         if(i < j)swap(y[i],y[j]);
         k = len/;
         while( j >= k) {
             j -= k;
             k /= ;
         }
         if(j < k) j += k;
     }
 }

 void fft(Complex y[],int len,int on) {
     change(y,len);
     for(int h = ; h <= len; h <<= ) {
         Complex wn(cos(-on**PI/h),sin(-on**PI/h));
         for(int j = ;j < len;j+=h) {
             Complex w(,);
             for(int k = j;k < j+h/;k++) {
                 Complex u = y[k];
                 Complex t = w*y[k+h/];
                 y[k] = u+t;
                 y[k+h/] = u-t;
                 w = w*wn;
             }
         }
     }
     if(on == -) {
         for(int i = ;i < len;i++) {
             y[i].r /= len;
         }
     }
 }

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