https://nanti.jisuanke.com/t/41400
对于大的情况,我们依次对a与b,a与c,c与b去卷积,然后答案就是$n^3$,卷积后减去不合法的情况
但是由于值域每次都是1e5,单组复杂度每次都是$1e5*log1e5$且常数很大
但是我们发现,只有20组n>1000
因此,对于小范围数据,我们直接使用$1000*1000$的暴力
#include<bits/stdc++.h> #define ll long long #define rep(ii,a,b) for(int ii=a;ii<=b;++ii) #define per(ii,a,b) for(int ii=b;ii>=a;--ii) using namespace std;//head const int maxn=4e5+10,maxm=2e6+10; const ll INF=0x3f3f3f3f,mod=1e9+7; int casn,n,m,k,kase=1,l,r; namespace fastio{//@支持读取整数,字符串,输出整数@ bool isdigit(char c){return c>=48&&c<=57;} const int maxsz=1e7; class fast_iostream{public: char ch=get_char(); bool endf=1,flag; char get_char(){ static char buffer[maxsz],*a=buffer,*b=buffer; return b==a&&(b=(a=buffer)+fread(buffer,1,maxsz, stdin),b==a)?EOF:*a++; } template<typename type>bool get_int(type& tmp){ flag=tmp=0; while(!isdigit(ch)&&ch!=EOF){flag=ch=='-';ch=get_char();}; if(ch==EOF)return endf=0; do{tmp=ch-48+tmp*10;}while(isdigit(ch=get_char())); if(flag)tmp=-tmp; return 1; } int get_str(char* str){ char* tmp=str; while(ch=='\r'||ch=='\n'||ch==' ')ch=get_char(); if(ch==EOF)return(endf=0),*tmp=0; do{*(tmp++)=ch;ch=get_char();}while(ch!='\r'&&ch!='\n'&&ch!=' '&&ch!=EOF); *(tmp++)=0; return(int)(tmp-str-1); } fast_iostream& operator>>(char* tmp){get_str(tmp);return *this;} template<typename type>fast_iostream& operator>>(type& tmp){get_int(tmp);return *this;} operator bool() const {return endf;} }; }fastio::fast_iostream io; const double pi=acos(-1.0); struct cp{double x,y;}; cp operator*(cp a,cp b){return {a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x};} cp operator+(cp a,cp b){return {a.x+b.x,a.y+b.y};} cp operator-(cp a,cp b){return {a.x-b.x,a.y-b.y};} class fourier{public: int rev[maxn],len,pw; void init(int n){ len=1,pw=0; while(len<=n) len<<=1,pw++; rep(i,0,len-1) rev[i]=0; rep(i,0,len-1) rev[i]=(rev[i>>1]>>1)|((i&1)<<(pw-1)); } void transform(cp*a,int flag){ rep(i,0,len-1) if(i<rev[i]) swap(a[i],a[rev[i]]); for(int mid=1;mid<len;mid<<=1){ cp wn={cos(pi/mid),flag*sin(pi/mid)}; for(int r=mid<<1,j=0;j<len;j+=r){ cp t={1,0}; for(int k=0;k<mid;k++,t=t*wn){ cp x=a[j+k],y=t*a[mid+j+k]; a[j+k]=x+y,a[j+k+mid]=x-y; } } } if(flag==-1) rep(i,0,len) a[i].x/=len; } }fft; int a[maxn],b[maxn],c[maxn]; cp cnta[maxn],cntb[maxn],cntc[maxn]; cp resa[maxn],resb[maxn],resc[maxn]; ll suma[maxn],sumb[maxn],sumc[maxn]; #define cin io int main(){ cin>>casn; while(kase<=casn){ cin>>n; fft.init(1e5+1e5+2); if(n<=1000){ rep(i,1,n)cin>>a[i]; rep(i,1,n)cin>>b[i]; rep(i,1,n)cin>>c[i]; ll ans=1ll*n*n*n; sort(c+1,c+1+n); sort(b+1,b+1+n); sort(a+1,a+1+n); rep(i,1,c[n]) suma[i]=0; rep(i,1,n)rep(j,1,n) ++suma[a[i]+b[j]]; rep(i,1,c[n])suma[i]+=suma[i-1]; rep(i,1,n) ans-=suma[c[i]-1]; rep(i,1,b[n]) suma[i]=0; rep(i,1,n)rep(j,1,n) ++suma[a[i]+c[j]]; rep(i,1,b[n])suma[i]+=suma[i-1]; rep(i,1,n) ans-=suma[b[i]-1]; rep(i,1,a[n]) suma[i]=0; rep(i,1,n)rep(j,1,n) ++suma[c[i]+b[j]]; rep(i,1,a[n])suma[i]+=suma[i-1]; rep(i,1,n) ans-=suma[a[i]-1]; printf("Case #%d: %lld\n",kase++,ans); continue; } rep(i,1,n) { cin>>a[i]; cnta[a[i]].x++; } rep(i,1,n) { cin>>b[i]; cntb[b[i]].x++; } rep(i,1,n) { cin>>c[i]; cntc[c[i]].x++; } fft.transform(cnta,1);fft.transform(cntb,1);fft.transform(cntc,1); rep(i,0,fft.len-1){ resa[i]=cntb[i]*cntc[i]; resb[i]=cnta[i]*cntc[i]; resc[i]=cntb[i]*cnta[i]; } fft.transform(resa,-1);fft.transform(resb,-1);fft.transform(resc,-1); rep(i,1,fft.len-1){ suma[i]=suma[i-1]+(ll)(resa[i].x+0.5); sumb[i]=sumb[i-1]+(ll)(resb[i].x+0.5); sumc[i]=sumc[i-1]+(ll)(resc[i].x+0.5); } ll ans=n*1ll*n*n; rep(i,1,n){ ans-=suma[a[i]-1]+sumb[b[i]-1]+sumc[c[i]-1]; } printf("Case #%d: %lld\n",kase++,ans); rep(i,0,fft.len){ cnta[i]=cntb[i]=cntc[i]=(cp){0,0}; } } } 来源：博客园
作者：nervending
链接：https://www.cnblogs.com/nervendnig/p/11525240.html