[HRBUSTOJ1476] Pairs (FFT)

Title Link: http://acm-software.hrbust.edu.cn/problem.php?id=1476

Test instructions: Give n number, M inquiry, ask a K each time. Ask the number of the n number of two and strictly less than k pairs.

According to the input sample, nothing is required:

f = CNT (1 2 3 4 5) T * (x) (where (x) represents a vector of 1,x,x^2 ..., CNT indicates the number of occurrences of these numbers, T means transpose) squared once corresponds to the coefficient before the X power, taking the first k-1 coefficient and is the answer. Complexity is too high, O (n^2) coefficient pointers is not possible, so need to make a point value expression DFT and then idft back.

1#include <bits/stdc++.h>2 using namespacestd;3 4 Const DoublePI = ACOs (-1.0);5typedefstructComplex {6     Doubler,i;7Complex (Double_r =0.0,Double_i =0.0) {8R = _r; i =_i;9     }TenComplexoperator+(ConstComplex &b) { One         returnComplex (r+b.r,i+B.I); A     } -Complexoperator-(ConstComplex &b) { -         returnComplex (r-b.r,i-B.I); the     } -Complexoperator*(ConstComplex &b) { -         returnComplex (r*b.r-i*b.i,r*b.i+i*B.R); -     } + }complex; -  + voidChange (Complex y[],intLen) { A     inti,j,k; at      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; in         } -         if(J < k) J + =K; to     } + } -  the voidFFT (Complex y[],intLenintOn ) { * Change (Y,len); $      for(inth =2; H <= Len; H <<=1) {Panax NotoginsengComplex wn (cos (-on*2*pi/h), sin (-on*2*pi/h)); -          for(intj =0; J < len;j+=h) { theComplex W (1,0); +              for(intK = J;k < j+h/2; k++) { AComplex U =Y[k]; theComplex T = w*y[k+h/2]; +Y[k] = u+T; -y[k+h/2] = uT; $W = w*WN; $             } -         } -     } the     if(On =-1) { -          for(inti =0; I < len;i++) {WuyiY[I].R/=Len; the         } -     } Wu } -  About Const intMAXN =400040; $typedefLong LongLL; - intA[MAXN]; - Complex C[MAXN]; - LL F[MAXN]; A intN, M; +  the intMain () { -     //freopen ("in", "R", stdin); $     intT; thescanf"%d", &T); the      while(t--) { theMemset (F,0,sizeof(f)); thescanf"%d%d",&n,&m); -         intMaxx =-1; in          for(inti =0; I < n; i++) { thescanf"%d", &a[i]); theMaxx =Max (A[i], Maxx); Aboutf[a[i]]++; the         } the         intLen1 = Maxx +1; the         intLen =1; +          while(Len <2* len1) Len <<=1; -          for(inti =0; i < Len; i++) C[i] = Complex (0,0); the          for(inti =0; i < len1; i++) C[i] = Complex (F[i],0);BayiFFT (c, Len,1); the          for(inti =0; i < Len; i++) C[i] = c[i] *C[i]; theFFT (c, Len,-1); -          for(inti =0; i < Len; i++) F[i] = (LL) (C[I].R +0.5); -Len =2*Maxx; the          for(inti =0; I < n; i++) f[a[i]*2]--; the          for(inti =1; I <= Len; i++) { theF[i]/=2; theF[i] + = f[i-1]; -         } the          while(m--) { the             intK; thescanf"%d", &k);94printf"%lld\n", f[k-1]); the         } the     } the     return 0;98}

[HRBUSTOJ1476] Pairs (FFT)