【bzoj3771】【xsy1728】Triple

【bzoj3771】【xsy1728】

题意

求\(\sum_{i}[a_i=S]+\sum_{i<j}[a_i+a_j=S]+\sum_{i<j<k}[a_i+a_j+a_k=S]\)

\(n\leq 30000\)，\(a_i\leq 40000\)

分析

容斥+FFT。

主要展现一些通过翻阅众多题解，得到的一些启示。

①把IDFT的除法放在内部。

void FFT(CP a[S],int len,int kd) {
    //...
    if (kd==-1) {
        rep(i,0,len-1)
            a[i].x/=len;
    }
}

②变量声明的合理放置

int lenA,lenB,lenR;
int bit,len; int rev[S];
struct CP {
    double x,y;
    CP (double _x=0,double _y=0) {
        x=_x,y=_y;
    }
    friend CP operator + (CP a,CP b) {
        return CP(a.x+b.x,a.y+b.y);
    }
    friend CP operator - (CP a,CP b) {
        return CP(a.x-b.x,a.y-b.y);
    }
    friend CP operator * (CP a,CP b) {
        return CP(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);
    }
}a[S],b[S],r[S],tmp[S];

③良好的封闭性Mut(a,lenA,b,lenB,r,lenR)

void Mut(CP a[S],int lenA,CP b[S],int lenB,CP r[S],int &lenR) {
    fill(r,r+S,0);
    lenR=lenA+lenB;

    len=1,bit=0;
    while (len<=lenA||len<=lenB)
        len<<=1,bit++;
    len<<=1,bit++;

    memset(rev,0,sizeof rev);
    rep(i,0,len-1) {
        int x=0;
        rep(j,0,bit-1)
            if (i>>j&1)
                x+=(1<<(bit-1-j));
        rev[i]=x;
    }

    FFT(a,len,1);
    FFT(b,len,1);
    rep(i,0,len-1) r[i]=a[i]*b[i];
    FFT(r,len,-1);
}
LL Trans(CP a) {
    return (LL)(a.x+0.5);
}

注意要使用Fill而不能使用memset。

④生成函数对数组的初始化

由于大量使用了相同的生成函数，所以写作一个过程。

int cmp(double x) {
    if (fabs(x)<EPS) return 0;
    return x<0?-1:1;
}
void IA(CP *a,int &lenA) {
    fill(a,a+S,0);
    rep(i,1,n) a[w[i]].x+=1.0;

    lenA=C1;
    while (lenA>=0&&!cmp(a[lenA].x))
        lenA--;
}
void IA2(CP a[S],int &lenA) {
    fill(a,a+S,0);
    rep(i,1,n) a[w[i]+w[i]].x+=1.0;

    lenA=C2;
    while (lenA>=0&&!cmp(a[lenA].x))
        lenA--;
}
void IA3(CP a[S],int &lenA) {
    fill(a,a+S,0);
    rep(i,1,n) a[w[i]+w[i]+w[i]].x+=1.0;

    lenA=C3;
    while (lenA>=0&&!cmp(a[lenA].x))
        lenA--;
}

⑤生成函数支持加法的合并

写了199行。

然后看了一下别人的代码。

woc77行......

发现生成函数支持加法的合并。

所以先用生成函数算出所有情况，然后再一次转换回来就好了。

代码

初次的AC代码：

#include <cstdio>
#include <cstring>
#include <cctype>
#include <cmath>
#include <algorithm>
using namespace std;

#define rep(i,a,b) for (int i=(a);i<=(b);i++)

typedef long long LL;

const int N=65536;
const int S=262144;
const double PI=M_PI;

const int C1=40000;
const int C2=80000;
const int C3=120000;

const double EPS=1e-8;

int n;
int w[N];

LL cnt[S];

LL c[S];
int lenA,lenB,lenR;

int bit,len; int rev[S];
struct CP {
    double x,y;
    CP (double _x=0,double _y=0) {
        x=_x,y=_y;
    }
    friend CP operator + (CP a,CP b) {
        return CP(a.x+b.x,a.y+b.y);
    }
    friend CP operator - (CP a,CP b) {
        return CP(a.x-b.x,a.y-b.y);
    }
    friend CP operator * (CP a,CP b) {
        return CP(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);
    }
}a[S],b[S],r[S],tmp[S];

int rd(void) {
    int x=0,f=1; char c=getchar();
    for (;!isdigit(c);c=getchar()) if (c=='-') f=-1;
    for (;isdigit(c);c=getchar()) x=x*10+c-'0';
    return x*f;
}

void FFT(CP a[S],int len,int kd) {
    rep(i,0,len-1) tmp[i]=a[rev[i]];
    rep(i,0,len-1) a[i]=tmp[i];

    for (int i=2;i<=len;i<<=1) {
        CP wn(cos(2*PI/i),sin(kd*2*PI/i));
        for (int k=0;k<len;k+=i) {
            CP w(1,0);
            rep(j,0,i/2-1) {
                CP x=a[k+j],y=w*a[k+j+i/2];
                a[k+j]=x+y,a[k+j+i/2]=x-y;
                w=w*wn;
            }
        }
    }

    if (kd==-1) {
        rep(i,0,len-1)
            a[i].x/=len;
    }
}
void Mut(CP a[S],int lenA,CP b[S],int lenB,CP r[S],int &lenR) {
    fill(r,r+S,0);
    lenR=lenA+lenB;

    len=1,bit=0;
    while (len<=lenA||len<=lenB)
        len<<=1,bit++;
    len<<=1,bit++;

    memset(rev,0,sizeof rev);
    rep(i,0,len-1) {
        int x=0;
        rep(j,0,bit-1)
            if (i>>j&1)
                x+=(1<<(bit-1-j));
        rev[i]=x;
    }

    FFT(a,len,1);
    FFT(b,len,1);
    rep(i,0,len-1) r[i]=a[i]*b[i];
    FFT(r,len,-1);
}
LL Trans(CP a) {
    return (LL)(a.x+0.5);
}

int cmp(double x) {
    if (fabs(x)<EPS) return 0;
    return x<0?-1:1;
}
void IA(CP *a,int &lenA) {
    fill(a,a+S,0);
    rep(i,1,n) a[w[i]].x+=1.0;

    lenA=C1;
    while (lenA>=0&&!cmp(a[lenA].x))
        lenA--;
}
void IA2(CP a[S],int &lenA) {
    fill(a,a+S,0);
    rep(i,1,n) a[w[i]+w[i]].x+=1.0;

    lenA=C2;
    while (lenA>=0&&!cmp(a[lenA].x))
        lenA--;
}
void IA3(CP a[S],int &lenA) {
    fill(a,a+S,0);
    rep(i,1,n) a[w[i]+w[i]+w[i]].x+=1.0;

    lenA=C3;
    while (lenA>=0&&!cmp(a[lenA].x))
        lenA--;
}

void Init(void) {
    n=rd();
    rep(i,1,n) w[i]=rd();
}
void Solve1(void) {
    IA(a,lenA);
    rep(i,1,lenA) cnt[i]+=Trans(a[i]);
}
void Solve2(void) {
    memset(c,0,sizeof c);

    IA(a,lenA); IA(b,lenB);
    Mut(a,lenA,b,lenB,r,lenR);
    rep(i,0,lenR) c[i]+=Trans(r[i]);

    IA2(a,lenA);
    rep(i,0,lenA) c[i]-=Trans(a[i]);

    rep(i,0,S-1)
        c[i]>>=1;

    rep(i,0,S-1)
        cnt[i]+=c[i];
}
void Solve3(void) {
    memset(c,0,sizeof c);

    // \sum(x,y,z)
    IA(a,lenA); IA(b,lenB);
    Mut(a,lenA,b,lenB,r,lenR);
    IA(a,lenA);
    Mut(a,lenA,r,lenR,b,lenB);
    rep(i,0,lenB) c[i]+=Trans(b[i]);;

    // \sum(x,x,y)
    IA2(a,lenA); IA(b,lenB);
    Mut(a,lenA,b,lenB,r,lenR);
    rep(i,0,lenR) c[i]-=3*Trans(r[i]);

    // \sum(x,x,x)
    IA3(a,lenA);
    rep(i,0,lenA) c[i]+=2*Trans(a[i]);

    rep(i,0,S-1)
        c[i]/=6;

    rep(i,0,S-1)
        cnt[i]+=c[i];
}
void Print(void) {
    rep(i,0,S-1)
        if (cnt[i]>0)
            printf("%d %lld\n",i,cnt[i]);
}

int main(void) {
    #ifndef ONLINE_JUDGE
    freopen("xsy1728.in","r",stdin);
    freopen("xsy1728.out","w",stdout);
    #endif

    Init();
    Solve1();
    Solve2();
    Solve3();
    Print();

    return 0;
}