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  • bzoj3527: [Zjoi2014]力 fft

    题意:求(E_i=sum_{j=1}^{i-1}qj/{(i-j)^2}-sum_{j=i+1}^{n}qj/{(i-j)^2})
    题解:构造前几个Ei,可以发现(E_i=a_i*b_{j-i}),(a_i=q_i),(b=-1/(n-1)^2-...-1/1^2-0+1/1^2+...+1/n^2)
    卷积搞一搞就行了

    /**************************************************************
        Problem: 3527
        User: walfy
        Language: C++
        Result: Accepted
        Time:2748 ms
        Memory:29420 kb
    ****************************************************************/
     
    //#pragma comment(linker, "/stack:200000000")
    //#pragma GCC optimize("Ofast,no-stack-protector")
    //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
    //#pragma GCC optimize("unroll-loops")
    #include<bits/stdc++.h>
    #define fi first
    #define se second
    #define db double
    #define mp make_pair
    #define pb push_back
    #define pi acos(-1.0)
    #define ll long long
    #define vi vector<int>
    #define mod 1000000007
    #define ld long double
    #define C 0.5772156649
    #define ls l,m,rt<<1
    #define rs m+1,r,rt<<1|1
    #define pll pair<ll,ll>
    #define pil pair<int,ll>
    #define pli pair<ll,int>
    #define pii pair<int,int>
    //#define cd complex<double>
    #define ull unsigned long long
    #define base 1000000000000000000
    #define Max(a,b) ((a)>(b)?(a):(b))
    #define Min(a,b) ((a)<(b)?(a):(b))
    #define fio ios::sync_with_stdio(false);cin.tie(0)
    template<typename T>
    inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
    template<typename T>
    inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
    inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
    inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
    inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
    inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
    inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}
     
    using namespace std;
     
    const double eps=1e-8;
    const ll INF=0x3f3f3f3f3f3f3f3f;
    const int N=100000+10,maxn=400000+10,inf=0x3f3f3f3f;
     
    struct cd{
        db x,y;
        cd(db _x=0.0,db _y=0.0):x(_x),y(_y){}
        cd operator +(const cd &b)const{
            return cd(x+b.x,y+b.y);
        }
        cd operator -(const cd &b)const{
            return cd(x-b.x,y-b.y);
        }
        cd operator *(const cd &b)const{
            return cd(x*b.x - y*b.y,x*b.y + y*b.x);
        }
        cd operator /(const db &b)const{
            return cd(x/b,y/b);
        }
    }a[N<<3],b[N<<3];
    int rev[N<<3];
    void getrev(int bit)
    {
        for(int i=0;i<(1<<bit);i++)
            rev[i]=(rev[i>>1]>>1) | ((i&1)<<(bit-1));
    }
    void fft(cd *a,int n,int dft)
    {
        for(int i=0;i<n;i++)
            if(i<rev[i])
                swap(a[i],a[rev[i]]);
        for(int step=1;step<n;step<<=1)
        {
            cd wn(cos(dft*pi/step),sin(dft*pi/step));
            for(int j=0;j<n;j+=step<<1)
            {
                cd wnk(1,0);
                for(int k=j;k<j+step;k++)
                {
                    cd x=a[k];
                    cd y=wnk*a[k+step];
                    a[k]=x+y;a[k+step]=x-y;
                    wnk=wnk*wn;
                }
            }
        }
        if(dft==-1)for(int i=0;i<n;i++)a[i]=a[i]/n;
    }
    int main()
    {
        int n;scanf("%d",&n);
        int sz=0;
        while((1<<sz)<n)sz++;
        sz++;getrev(sz);
        for(int i=0;i<=(1<<sz);i++)a[i]=b[i]=0;
        for(int i=0;i<n;i++)
        {
            double x;scanf("%lf",&x);
            a[i]=x;
        }
        int now=0;
        for(int i=n-1;i>=1;i--)b[now++]=-1.0/i/i;
        now++;
        for(int i=1;i<=n-1;i++)b[now++]=1.0/i/i;
        fft(a,(1<<sz),1);fft(b,(1<<sz),1);
        for(int i=0;i<=(1<<sz);i++)a[i]=a[i]*b[i];
        fft(a,(1<<sz),-1);
        for(int i=n-1;i<2*n-1;i++)printf("%lf
    ",a[i].x);
        return 0;
    }
    /********************
     
    ********************/
    
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  • 原文地址:https://www.cnblogs.com/acjiumeng/p/9440840.html
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