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hihoCoder1388 Periodic Signal(2016北京网赛F:NTT)
题目
Source
http://hihocoder.com/problemset/problem/1388
Description
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.
Input
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.
Output
For each test case, print the answer in a single line.
Sample Input
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
Sample Output
80
0
分析
题目大概说k多少时上面的式子值最小。
展开那个式子可以得到,$\sum A_i^2 + \sum B_i^2 - 2\sum A_iB_{i+k}$。前面是固定的,问题相当于最小化后面那一部分。
容易发现这是个可以用FFT快速求出的卷积形式。。
把A加长一倍,B反转,构造多项式,这样多项式乘积指数[n,2n)的系数就是各个位置的结果了。
直接FFT是不行的,千百亿的浮点数运算精度不够,比赛时搞了30多发没搞出来。。
我用了个NTT的板子A了,http://blog.csdn.net/u013654696/article/details/48653057,里面除了附赠了几个模数外还有一个O(1)快速乘不明觉厉。
代码
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define MAXN 262144const long long P=50000000001507329LL; // 190734863287 * 2 ^ 18 + 1//const int P=1004535809; // 479 * 2 ^ 21 + 1//const int P=998244353; // 119 * 2 ^ 23 + 1const int G=3;long long mul(long long x,long long y){ return (x*y-(long long)(x/(long double)P*y+1e-3)*P+P)%P;}long long qpow(long long x,long long k,long long p){ long long ret=1; while(k){ if(k&1) ret=mul(ret,x); k>>=1; x=mul(x,x); } return ret;}long long wn[25];void getwn(){ for(int i=1; i<=18; ++i){ int t=1<<i; wn[i]=qpow(G,(P-1)/t,P); }}int len;void NTT(long long y[],int op){ for(int i=1,j=len>>1,k; i<len-1; ++i){ if(i<j) swap(y[i],y[j]); k=len>>1; while(j>=k){ j-=k; k>>=1; } if(j<k) j+=k; } int id=0; for(int h=2; h<=len; h<<=1) { ++id; for(int i=0; i<len; i+=h){ long long w=1; for(int j=i; j<i+(h>>1); ++j){ long long u=y[j],t=mul(y[j+h/2],w); y[j]=u+t; if(y[j]>=P) y[j]-=P; y[j+h/2]=u-t+P; if(y[j+h/2]>=P) y[j+h/2]-=P; w=mul(w,wn[id]); } } } if(op==-1){ for(int i=1; i<len/2; ++i) swap(y[i],y[len-i]); long long inv=qpow(len,P-2,P); for(int i=0; i<len; ++i) y[i]=mul(y[i],inv); }}void Convolution(long long A[],long long B[],int n){ for(len=1; len<(n<<1); len<<=1); for(int i=n; i<len; ++i){ A[i]=B[i]=0; } NTT(A,1); NTT(B,1); for(int i=0; i<len; ++i){ A[i]=mul(A[i],B[i]); } NTT(A,-1);}long long A[MAXN],B[MAXN];int main(){ getwn(); int t,n; scanf("%d",&t); while(t--){ scanf("%d",&n); long long ans=0; for(int i=0; i<n; ++i){ scanf("%lld",&A[i]); ans+=A[i]*A[i]; } for(int i=0; i<n; ++i){ scanf("%lld",&B[n-i-1]); ans+=B[n-i-1]*B[n-i-1]; } for(int i=0; i<n; ++i){ A[i+n]=A[i]; B[i+n]=0; } Convolution(A,B,2*n); long long mx=0; for(int i=n; i<2*n; ++i){ mx=max(mx,A[i]); } printf("%lld\n",ans-2*mx); } return 0;}
hihoCoder1388 Periodic Signal(2016北京网赛F:NTT)