题目链接:http://hihocoder.com/problemset/problem/1388?sid=974337
题目大意:找出一个$k$,使得$sum_{i=0}^{n-1}(A_{i}-B_{i+k quad mod quad n})^{2}$最小
把那个式子拆开..得到:
$sum _{i=0}^{n-1} A_{i}^{2}+sum_{i=0}^{n-1}B_{i}^{2}-sum_{i=0}^{n-1}2*A_{i}*B_{i+k}$
前面的两个和式是定值,所以最小化后面的值,后面这个是个卷积的形式,所以把$B$翻转硬上就好了...
然后问题就来了..FFT了之后貌似精度爆炸了..然后就写NTT..然后又炸了..发现模数规模不够..于是查表换了个巨大的模数,然后又炸了..
然后去看题解,意识到一个严重的问题..相乘爆longlong...然后又学习了TA爷的$O(1)$快速乘..
附上一张神表:折跃
#include<iostream> #include<cstdio> #include<algorithm> #include<cmath> #include<cstring> using namespace std; #define LL long long inline LL read() { LL x=0,f=1; char ch=getchar(); while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();} while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();} return x*f; } #define P (LL)((29LL<<57)+1) #define G 3 #define MAXN 800100 int T,N,len; LL Wn[30],A[MAXN],B[MAXN],C[MAXN],wn[MAXN],a[MAXN],b[MAXN]; inline LL Mul(LL x,LL y) {return (x*y-(LL)(x/(long double)P*y+1e-3)*P+P)%P;} inline LL Pow(LL x,LL y) { LL re=1; for (LL i=y; i; i>>=1,x=Mul(x,x)) if (i&1) re=Mul(re,x); return re; } inline LL Inv(LL x) {return Pow(x,P-2);} inline void Prework() { len=1; while (len<(N<<1)) len<<=1; for (int i=0; i<N; i++) A[i]=a[i]; for (int i=0; i<N; i++) B[i]=b[N-1-i]; for (int i=N; i<len; i++) A[i]=0; for (int i=N; i<len; i++) B[i]=0; // for (int i=0; i<len; i++) printf("%I64d ",A[i]); puts(""); // for (int i=0; i<len; i++) printf("%I64d ",B[i]); puts(""); for (int i=0; i<=28; i++) wn[i]=Pow(G,(P-1)/(1<<i)); } inline void Rader(LL *x) { for (int i=1,j=len>>1,k; i<len-1; i++) { if (i<j) swap(x[i],x[j]); k=len>>1; while (j>=k) j-=k,k>>=1; if (j<k) j+=k; } } inline void DFT(LL *x,int opt) { Rader(x); for (int h=2,id=0; h<=len; h<<=1) { LL Wn=wn[++id]; for (int i=0; i<len; i+=h) { LL W=1; for (int j=i; j<i+h/2; j++) { LL u=x[j]%P,t=Mul(W,x[j+h/2]); x[j]=(u+t)%P; x[j+h/2]=(u-t+P)%P; W=Mul(W,Wn); } } } if (opt==-1) { for (int i=1; i<len/2; i++) swap(x[i],x[len-i]); for (int i=0; i<len; i++) x[i]=Mul(x[i],Inv(len)); } } inline void NTT(LL *A,LL *B) { DFT(A,1); DFT(B,1); for (int i=0; i<len; i++) C[i]=Mul(A[i],B[i]); DFT(C,-1); } int main() { T=read(); while (T--) { N=read(); LL ans=0; for (int i=0; i<=N-1; i++) a[i]=read(); for (int i=0; i<=N-1; i++) b[i]=read(); for (int i=0; i<=N-1; i++) ans+=(LL)a[i]*a[i]; for (int i=0; i<=N-1; i++) ans+=(LL)b[i]*b[i]; Prework(); NTT(A,B); LL mx=C[N-1]; for (int i=0; i<N-1; i++) mx=max(mx,C[i]+C[i+N]); // for (int i=0; i<len; i++) printf("%I64d ",C[i]); puts(""); printf("%lld ",ans-2*mx); } return 0; }