礼物 HYSBZ - 4827 (fft + 构造 )

 1 #include"bits/stdc++.h"
 2 #define sd(x) scanf("%d",&(x));
 3 #define sf(x) scanf("%lf",&(x));
 4 #define sld(x) scanf("%lld",&(x));
 5 using namespace std;
 6 
 7 const int maxn = 2e6+10;
 8 const double Pi = acos(-1.0);
 9 struct cp
10 {
11     double x,y;
12     cp (double xx=0,double yy=0)
13     {x=xx,y=yy;}
14 }a[maxn],g[maxn],b[maxn];
15 
16 cp operator + (cp a,cp b){ return cp(a.x+b.x , a.y+b.y);}
17 cp operator - (cp a,cp b){ return cp(a.x-b.x , a.y-b.y);}
18 cp operator * (cp a,cp b){ return cp(a.x*b.x-a.y*b.y , a.x*b.y+a.y*b.x);}//不懂的看复数的运算那部分
19 
20 int n,m;
21 int l,r[maxn];
22 int limit = 1;
23 int s1,s2,f1,f2;
24 
25 
26 inline void fft(cp *a,int ff)
27 {
28     for(int i=0;i<limit;i++)
29     if(i<r[i])swap(a[i],a[r[i]]);
30     for(int mid=1;mid<limit;mid<<=1)
31         {
32         cp wn(cos(Pi/mid) , ff*sin(Pi/mid));
33         for(int R=mid<<1,j=0;j<limit;j+=R)
34         {
35             cp w(1,0);
36             for(int k=0;k<mid;k++,w=w*wn)
37             {
38                 cp x=a[j+k],y=w*a[j+mid+k];
39                 a[j+k]=x+y;
40                 a[j+mid+k]=x-y;
41             }
42         }
43 
44     }
45 
46 }
47 
48 int main()
49 {
50 
51    sd(n); sd(m);
52    for(int i=0;i<n;i++)sf(a[n-1-i].x);
53    for(int i=0;i<n;i++)sf(b[i].x);
54    for(int i=0;i<n;i++)b[n+i]=b[i];
55 
56   for(int i=0;i<n;i++)
57     f1+=(int)a[i].x*(int)a[i].x,
58     f2+=(int)b[i].x*(int)b[i].x,
59     s1+=(int)a[i].x,
60     s2+=(int)b[i].x;
61 
62    while(limit<=3*n) limit<<=1,l++;//最高次项为 3*n
63   // cout<<"LIMIT: "<<limit<<endl;
64    for(int i=0;i<limit;i++)
65    r[i]=(r[i>>1]>>1)|( (i&1)<<(l-1));
66 
67 
68    fft(a,1); fft(b,1);
69    for(int i=0;i<=limit;i++)a[i]=a[i]*b[i];
70    fft(a,-1);
71    int ans = 2e9;
72 
73 
74 
75    for(int i=0;i<limit;i++)a[i].x=(int)(a[i].x/limit+0.5);
76   // cout<<f1<<" "<<f2<<" "<<s1<<" "<<s2<<endl;
77    for(int i=0;i<n;i++)
78    {
79        for(int c=0;c<m;c++)
80        {
81            ans=min(ans,f1+f2-2*(int)a[n+i].x+2*c*(s1-s2)+n*c*c);
82        }
83    }
84    cout<<ans;
85 
86 
87    return 0;
88 
89 
90 
91 
92 }