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Profess X is an expert in signal processing. He has a device which can send a particular 1 second signal repeatedly. The signal is A0 ... An-1 under n Hz sampling.
One day, the device fell on the ground accidentally. Profess X wanted to check whether the device can still work properly. So he ran another n Hz sampling to the fallen device and got B0 ... Bn-1.
To compare two periodic signals, Profess X define the DIFFERENCE of signal A and B as follow:
You may assume that two signals are the same if their DIFFERENCE is small enough.
Profess X is too busy to calculate this value. So the calculation is on you.
The first line contains a single integer T, indicating the number of test cases.
In each test case, the first line contains an integer n. The second line contains n integers, A0 ... An-1. The third line contains n integers, B0 ... Bn-1.
T≤40 including several small test cases and no more than 4 large test cases.
For small test cases, 0<n≤6⋅103.
For large test cases, 0<n≤6⋅104.
For all test cases, 0≤Ai,Bi<220.
For each test case, print the answer in a single line.
2 9 3 0 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 5 1 2 3 4 5 2 3 4 5 1
80 0
/* * hihocoder 1388 Periodic Signal * * 把式子变形一下就是求Ai*Bi+k求和的最大值,想到用FFT来求 * 会因为精度问题不能过,这是找出最小的那个k,然后重新算即可。 */ #include <bits/stdc++.h> using namespace std; const int MAXN = 200000+10; const double PI = acos(-1.0); struct Complex { double r,i; Complex(double _r=0,double _i=0):r(_r),i(_i){} Complex operator + (const Complex& rhs) { return Complex(r+rhs.r,i+rhs.i); } Complex operator - (const Complex& rhs) { return Complex(r-rhs.r,i-rhs.i); } Complex operator * (const Complex &rhs) { return Complex(r*rhs.r - i*rhs.i,i*rhs.r + r*rhs.i); } }; /* * 进行FFT和IFFT前的反转变换。 * 位置i和 (i二进制反转后位置)互换 * len必须取2的幂 */ void Rader(Complex F[],int len) { int j = len >> 1; for(int i = 1;i < len - 1;++i) { if(i < j) swap(F[i],F[j]); // reverse int k = len>>1; while(j>=k) { j -= k; k >>= 1; } if(j < k) j += k; } } /* * 做FFT * len必须为2^k形式, * on==1时是DFT,on==-1时是IDFT */ void FFT(Complex F[],int len,int t) { Rader(F,len); for(int h=2;h<=len;h<<=1) { Complex wn(cos(-t*2*PI/h),sin(-t*2*PI/h)); for(int j=0;j<len;j+=h) { Complex E(1,0); //旋转因子 for(int k=j;k<j+h/2;++k) { Complex u = F[k]; Complex v = E*F[k+h/2]; F[k] = u+v; F[k+h/2] = u-v; E=E*wn; } } } if(t==-1) //IDFT for(int i=0;i<len;++i) F[i].r/=len; } void Conv(Complex a[],Complex b[],int len) //求卷积 { FFT(a,len,1); FFT(b,len,1); for(int i=0;i<len;++i) a[i] = a[i]*b[i]; FFT(a,len,-1); } Complex va[MAXN],vb[MAXN]; int A[MAXN],B[MAXN],n,len; int AA[MAXN]; void init() { scanf("%d",&n); for(int i=0;i<n;i++) scanf("%d",&A[i]); for(int i=0;i<n;i++) scanf("%d",&B[i]); for(int i=n;i<2*n;i++) B[i]=B[i-n]; for(int i=0;i<n;i++) AA[i]=A[n-i-1]; len=1; while(len<2*n) len<<=1; for(int i=0;i<n;i++) va[i]=Complex(AA[i],0); for(int i=n;i<len;i++) va[i]=Complex(0,0); for(int i=0;i<2*n;i++) vb[i]=Complex(B[i],0); for(int i=2*n;i<len;i++) vb[i]=Complex(0,0); } void solve() { init(); Conv(va,vb,len); double ans=0; int k; for(int i=n-1;i<=2*n-2;i++) { if(va[i].r>ans) { ans=va[i].r; k=i; } } k-=n-1; long long a=0,b=0,ab=0; for(int i=0;i<n;i++) a+=(long long)A[i]*A[i]; for(int i=0;i<n;i++) b+=(long long)B[i]*B[i]; for(int i=0;i<n;i++) ab+=(long long)A[i]*B[i+k]; long long res=a+b-2*ab; printf("%lld\n",res); } int main() { int T; int n; scanf("%d",&T); while(T--) { solve(); } return 0; }
hihocoder 1388 Periodic Signal
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原文地址:http://www.cnblogs.com/wangdongkai/p/5906954.html