[玲珑OJ1044] Quailty and Binary Operation (FFT+cdq分治)

题目链接

题意：给定两个长度为n的数组a与长度为m的数组b, 给定一个操作符op满足 x op y = x < y ? x+y : x-y.  有q个询问，每次给出询问c，问：有多少对(i, j)满足a[i] op b[j] = c ？

0 <= c <= 100000, 其余数据范围在[0, 50000].

题解：问题的关键在于如何分隔开 x < y与x >= y. cdq分治，合并的时候a[l, mid]与b[mid+1, r]卷积一次计算a[] < b[] ， a[mid+1, r]与b[l, mid]再卷积一次a[] > b[]即可。

卡时，memset的时候优化了一下。

 #include <bits/stdc++.h>
 typedef long long ll;
 using namespace std;
 const int N = 1e5+;
 struct comp{
     double r,i;comp(double _r=,double _i=){r=_r;i=_i;}
     comp operator+(const comp x){return comp(r+x.r,i+x.i);}
     comp operator-(const comp x){return comp(r-x.r,i-x.i);}
     comp operator*(const comp x){return comp(r*x.r-i*x.i,r*x.i+i*x.r);}
 }x[N<<], y[N<<];
 const double pi=acos(-1.0);
 void FFT(comp a[],int n,int t){
     for(int i=,j=;i<n-;i++){
         for(int s=n;j^=s>>=,~j&s;);
         if(i<j)swap(a[i],a[j]);
     }
     for(int d=;(<<d)<n;d++){
         int m=<<d,m2=m<<;
         double o=pi/m*t;comp _w(cos(o),sin(o));
         for(int i=;i<n;i+=m2){
             comp w(,);
             for(int j=;j<m;j++){
                 comp &A=a[i+j+m],&B=a[i+j],t=w*A;
                 A=B-t;B=B+t;w=w*_w;
             }
         }
     }
     if(t==-)for(int i=;i<n;i++)a[i].r/=n;
 }
 int a[N], b[N], n, m, q;
 ll ans[N];
 void cdq(int l, int r){
     if(l == r){
         ans[] += a[l]*b[l];
         return;
     }
     int mid = l+r >> ;
     cdq(l, mid);
     int len = ;
     while(len <= (r-l+)) len <<= ;
     memset(x, , sizeof(comp)*len );
     memset(y, , sizeof(comp)*len );
     for(int i = l; i <= mid; i++)
         x[i-l] = comp(a[i], );
     for(int i = mid+; i <= r; i++)
         y[i-mid-] = comp(b[i], );
     FFT(x, len, ); FFT(y, len, );
     for(int i = ; i < len; i++)
         x[i] = x[i]*y[i];
     FFT(x, len, -);
     for(int i = l+mid+; i <= mid+r; i++)
         ans[i] += x[i-l-mid-].r+0.5;
     for(int i = ; i < len; i++)
         x[i] = y[i] = comp(, );
     for(int i = mid+; i <= r; i++)
         x[i-mid-] = comp(a[i], );
     for(int i = l; i <= mid; i++)
         y[mid+-i] = comp(b[i], );
     FFT(x, len, ); FFT(y, len, );
     for(int i = ; i < len; i++)
         x[i] = x[i]*y[i];
     FFT(x, len, -);
     for(int i = ; i <= r-l; i++)
         ans[i] += x[i].r+0.5;
     cdq(mid+, r);
 }

 int main(){
     int t, x, maxn; scanf("%d", &t);
     while(t--){
         scanf("%d%d%d", &n, &m, &q);
         maxn = ;
         for(int i = ; i < n; i++){
             scanf("%d", &x);
             maxn = max(maxn, x);
             a[x]++;
         }
         for(int i = ; i < m; i++){
             scanf("%d", &x);
             maxn = max(maxn, x);
             b[x]++;
         }
         cdq(, maxn);
         while(q--){
             scanf("%d", &x);
             printf("%lld\n", ans[x]);
         }
         memset(a, , sizeof(int)*(maxn+));
         memset(b, , sizeof(int)*(maxn+));
         memset(ans, , sizeof(ll)*(maxn*+));
     }
     return ;
 }