BZOJ4836 [Lydsy2017年4月月赛]二元运算

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如果只是求x+y=c的对数可以直接fft预处理每个c的答案然后O(1)出解

求x-y=c就把底下那个翻转一下然后fft

但是他要求在x<y的时候是x+y=c，x>=y的时候是x-y=c

还是考虑预处理每个c的答案然后O(1)出解

那么我们就考虑分治，对于权值区间[l,r]，对x在[l,mid]，y在(mid,r]里的做一次fft，对x在(mid,r]，y在[l,mid]里的做一次fft，这样就算出了跨mid的所有数对对答案的影响，然后递归[l,mid]和(mid,r)即可

复杂度T*mx log^2 mx，mx为最大权值

用了对实序列的FFT优化和读入优化才卡进去，时限8s跑了8900ms……但是感觉复杂度上不太能更优了？（可能是我太愚蠢了）应该是有常数更小的做法把

#include<iostream>#include<cstring>#include<ctime>#include<cmath>#include<algorithm>#include<iomanip>#include<cstdlib>#include<cstdio>#include<map>#include<bitset>#include<set>#include<stack>#include<vector>#include<queue>using namespace std;#define MAXN 200010#define MAXM 1010#define ll long long#define eps 1e-8#define MOD 1000000007#define INF 1000000000char xB[1<<15],*xS=xB,*xTT=xB;#define getc() (xS==xTT&&(xTT=(xS=xB)+fread(xB,1,1<<15,stdin),xS==xTT)?0:*xS++)#define isd(c) (c>='0'&&c<='9')int read(){char xchh;int xaa;    while(xchh=getc(),!isd(xchh));(xaa=xchh-'0');    while(xchh=getc(),isd(xchh))xaa=xaa*10+xchh-'0';return xaa;}const double pi=acos(-1);struct cl{double x;double y;cl(){}cl(double _x,double _y){x=_x;y=_y;}friend cl operator +(const cl &x,const cl &y){return cl(x.x+y.x,x.y+y.y);}friend cl operator -(const cl &x,const cl &y){return cl(x.x-y.x,x.y-y.y);}friend cl operator *(const cl &x,const cl &y){return cl(x.x*y.x-x.y*y.y,x.x*y.y+x.y*y.x);}friend cl operator /(const cl &x,const double &y){return cl(x.x/y,x.y/y);}cl con(){return cl(x,-y);}};cl a[MAXN],b[MAXN],c[MAXN];int A[MAXN],B[MAXN];ll ans[MAXN];int n1,n2,q;int N;int R[MAXN];void fft(cl *a,int f,int n,int L){int i,j,k;for(i=0;i<n;i++){R[i]=R[i>>1]>>1|((i&1)<<(L-1));if(i<R[i]){swap(a[i],a[R[i]]);}}for(i=1;i<n;i<<=1){cl wn(cos(pi/i),f*sin(pi/i));for(j=0;j<n;j+=i<<1){cl w(1,0);for(k=0;k<i;k++,w=w*wn){cl x=a[j+k],y=w*a[j+i+k];a[j+k]=x+y;a[j+i+k]=x-y;}}}if(f==-1){for(i=0;i<n;i++){a[i]=a[i]/n;}}}void cal(int l,int r,int n,int L){if(l==r){ans[0]+=(ll)A[l]*B[l];return ;}int i,j;int mid=l+r>>1;for(i=0;i<n;i++){a[i]=b[i]=cl(0,0);}for(i=l;i<=mid;i++){a[i-l].x=A[i];}for(i=mid+1;i<=r;i++){a[i-(mid+1)].y=B[i];}fft(a,1,n,L);    for(i=0;i<n;i++){        j=(n-i)&(n-1);        b[i]=(a[i]*a[i]-(a[j]*a[j]).con())*cl(0,-.25);    }    fft(b,-1,n,L);for(i=0;i<n;i++){ans[l+mid+1+i]+=(ll)(b[i].x+.1);}for(i=0;i<n;i++){a[i]=b[i]=cl(0,0);}for(i=mid+1;i<=r;i++){a[i-(mid+1)].x=A[i];}for(i=l;i<=mid;i++){a[mid-i].y=B[i];}fft(a,1,n,L);    for(i=0;i<n;i++){        j=(n-i)&(n-1);        b[i]=(a[i]*a[i]-(a[j]*a[j]).con())*cl(0,-.25);    }    fft(b,-1,n,L);for(i=0;i<n;i++){ans[i+1]+=(ll)(b[i].x+.1);}cal(l,mid,n>>1,L-1);cal(mid+1,r,n>>1,L-1);}int main(){int i,x;int tmp;tmp=read();while(tmp--){memset(A,0,sizeof(A));memset(B,0,sizeof(B));memset(ans,0,sizeof(ans));n1=read();n2=read();q=read();int mx=0;for(i=1;i<=n1;i++){x=read();A[x]++;mx=max(mx,x);}for(i=1;i<=n2;i++){x=read();B[x]++;mx=max(mx,x);}int n,L;for(L=0,n=1;n<mx<<1;n<<=1){L++;}cal(0,n-1,n,L);while(q--){x=read();printf("%lld\n",ans[x]);}}return 0;}/*11 1 1134*/

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