BZOJ 3771 Triple

研究母函数&容斥

刚开始没发现是不重复感觉大概很麻烦...

这篇题解不错

//By Richard#include <cstdio>#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <cmath>#include <ctime>#define rep(x,y,z) for (int x=(y);(x)<=(z);(x)++)#define per(x,y,z) for (int x=(y);(x)>=(z);(x)--)#define log2(x) (31-__builtin_clz(x))#define mod (int)(1e9+7)#define inf 0x3f3f3f3f#define cls(x) memset(x,0,sizeof(x))#ifdef DEBUG#define debugdo(X) X#define debugndo(X)#define debugout(X) cout<<(#X)<<"="<<(X)<<endl#else#define debugdo(X)#define debugndo(X) X#define debugout(X)#endif // debug#ifdef ONLINE_JUDGE#define debugdo(X)#define debugndo(X)#define debugout(X)#endif#define putarray(x,n) rep(iiii,1,n) printf("%d ",x[iiii])#define mp make_pairusing namespace std;typedef pair<int,int> pairs;typedef long long LL;/////////////////////read3.0////////////////////////////////////template <typename T>inline void read(T &x){char ch;x=0;bool flag=false;ch=getchar();while (ch>'9'||ch<'0') {ch=getchar();if (ch=='-') flag=true;}while ((ch<='9'&&ch>='0')){x=x*10+ch-'0';ch=getchar();}if (flag) x*=-1;}template <typename T>inline void read(T &x,T &y){read(x);read(y);}/////////////////variables&functions////////////////////const int maxn=1048576;const double pi=3.1415926535897932384626433832795028841971693993751058209749446,eps=0.1;int n,m,r[maxn],l,ans[maxn],N,x,inp[maxn];struct cp{double x,y;cp(double a=0,double b=0):x(a),y(b){}cp operator+(cp b){return cp(x+b.x,y+b.y);}cp operator-(cp b){return cp(x-b.x,y-b.y);}cp operator*(cp b){return cp(x*b.x-y*b.y,x*b.y+y*b.x);}cp operator/(int b){return cp(x/b,x/b);}}A[maxn],B[maxn],C[maxn];inline void swap(cp &a,cp &b){cp t=a;a=b;b=t;}void FFT(cp *a,int n,int op){rep(i,0,n-1) if (i<r[i]) swap(a[i],a[r[i]]);for (int i=1;i<n;i<<=1){cp wn(cos(pi/i),sin(pi/i)*op);for (int j=0;j<n;j+=i<<1){cp w(1,0);for (int k=0;k<i;k++,w=w*wn){cp x=a[j+k],y=a[i+j+k]*w;a[j+k]=x+y;a[i+j+k]=x-y;}}}if (op==-1) rep(i,0,n-1) a[i]=a[i]/n;}void init(){m+=n-2;for (n=1,l=0;n<=m;n<<=1,l++);rep(i,0,n-1) r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));}void mult(cp *a,cp *b){FFT(a,n,1);FFT(b,n,1);rep(i,0,n-1) a[i]=a[i]*b[i];FFT(a,n,-1);// debugdo(rep(i,0,m) printf("%d ",int(a[i].x+0.2)));}int main(){// read(n,m);// rep(i,0,n-1) read(A[i].x);// rep(i,0,m-1) read(B[i].x);// init();mult(A,B,true);read(N);int maxx=0;rep(i,0,N-1) {read(x);maxx=max(x,maxx);ans[x]++;A[x].x++;B[x].x++;inp[x]++;}n=maxx+1;m=maxx+1;init();mult(A,B);rep(i,0,m) ans[i]+=(int)((A[i].x-((i&1)?0:inp[i/2]))/2+eps);n=m+1;m=maxx+1;cls(B);rep(i,0,m-1) B[i].x=inp[i];rep(i,0,n) A[i].y=0;init();mult(A,B);int mm=m;for (int i=0;i<=maxx;i++) C[i<<1].x=inp[i];n=maxx*2+1;m=maxx+1;cls(B);rep(i,0,m-1) B[i].x=inp[i];init();mult(C,B);mm=max(mm,m);rep(i,0,mm) ans[i]+=(int)((A[i].x-3*C[i].x+2*((i%3)?0:inp[i/3]))/6+eps);rep(i,0,mm) if (ans[i]) printf("%d %d\n",i,ans[i]); return 0;}