傳送門：hihocoder #1388 : Periodic Signal?

先來幾個(gè)大牛傳送門：??(模板) NTT long long 版

解法一：因?yàn)槲覀冎繤FT會(huì)精度不夠，所以堅(jiān)持用NTT，但是模數(shù)不夠大，然后就一直GG，看來我們的搜索姿勢(shì)也有問題，居然沒有搜到上面大神的板子，真的是GG

http://www.cnblogs.com/WABoss/p/5903927.html

/**************************************************************Problem:User: youmiLanguage: C++Result: AcceptedTime:Memory: ****************************************************************/ //#pragma comment(linker, "/STACK:1024000000,1024000000") //#include<bits/stdc++.h> #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <map> #include <stack> #include <set> #include <sstream> #include <cmath> #include <queue> #include <deque> #include <string> #include <vector> #define zeros(a) memset(a,0,sizeof(a)) #define ones(a) memset(a,-1,sizeof(a)) #define sc(a) scanf("%d",&a) #define sc2(a,b) scanf("%d%d",&a,&b) #define sc3(a,b,c) scanf("%d%d%d",&a,&b,&c) #define scs(a) scanf("%s",a) #define sclld(a) scanf("%I64d",&a) #define pt(a) printf("%d\n",a) #define ptlld(a) printf("%I64d\n",a) #define rep(i,from,to) for(int i=from;i<=to;i++) #define irep(i,to,from) for(int i=to;i>=from;i--) #define Max(a,b) ((a)>(b)?(a):(b)) #define Min(a,b) ((a)<(b)?(a):(b)) #define lson (step<<1) #define rson (lson+1) #define eps 1e-6 #define oo 0x3fffffff #define TEST cout<<"*************************"<<endl const double pi=4*atan(1.0);using namespace std; typedef long long ll;int n; const ll P = 50000000001507329LL; //190734863287 * 2 ^ 18 + 1 //const ll P = 1004535809LL; //479 * 2 ^ 21 + 1 //const ll P = 1004535809; // 119 * 2 ^ 23 + 1 const int N = 1 << 18; const int G = 3; int len; ll A[N],B[N]; long long a[N],b[N],wn[30];ll mul(ll x, ll y) {return (x * y - (ll)(x / (long double)P * y + 1e-3) * P + P) % P; }ll qpow(ll x, ll k, ll p) {ll ret = 1;while(k) {if(k & 1) ret = mul(ret, x);k >>= 1;x = mul(x, x);}return ret; }void getwn() {for(int i = 1; i <= 18; ++i){int t = 1 << i;wn[i] = qpow(G, (P - 1) / t, P);} }void change(ll *y, int len) {for(int i = 1, j = len / 2; i < len - 1; ++i){if(i < j) swap(y[i], y[j]);int k = len / 2;while(j >= k){j -= k;k /= 2;}j += k;} }void NTT(ll *y, int len, int on) {change(y, len);int id = 0;for(int h = 2; h <= len; h <<= 1){++id;for(int j = 0; j < len; j += h){ll w = 1;for(int k = j; k < j + h / 2; ++k){ll u = y[k];ll t = mul(y[k+h/2], w);y[k] = u + t;if(y[k] >= P) y[k] -= P;y[k+h/2] = u - t + P;if(y[k+h/2] >= P) y[k+h/2] -= P;w = mul(w, wn[id]);}}}if(on == -1){for(int i = 1; i < len / 2; ++i) swap(y[i], y[len-i]);ll inv = qpow(len, P - 2, P);for(int i = 0; i < len; ++i)y[i] = mul(y[i], inv);} } void work()///卷積，點(diǎn)乘，插值 {NTT(a,len,1);NTT(b,len,1);for(int i=0;i<len;i++)a[i]=mul(a[i],b[i]);NTT(a,len,-1); } ll solve() {zeros(a);zeros(b);rep(i,0,n-1)a[i]=A[i];rep(i,0,n-1)b[i]=B[i];reverse(b,b+n);work();ll ans=0;rep(i,0,n-1)a[i]+=a[i+n];rep(i,0,n-1)ans=max(ans,2*a[i]);return ans; } int main() {#ifndef ONLINE_JUDGEfreopen("in.txt","r",stdin);#endifint T_T;scanf("%d",&T_T);getwn();for(int kase=1;kase<=T_T;kase++){sc(n);len=1;while(len<=2*n)len<<=1;rep(i,0,n-1)cin>>A[i];rep(i,0,n-1)cin>>B[i];ll temp=0;rep(i,0,n-1)temp+=A[i]*A[i];rep(i,0,n-1)temp+=B[i]*B[i];ll ans=solve();ans=temp-ans;cout<<(ans)<<endl;} } View Code

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解法二：這個(gè)解法確實(shí)很漂亮，比賽的時(shí)候一直徘徊找一個(gè)更大的 模數(shù)，然后就GG了，http://www.cnblogs.com/smartweed/p/5903838.html

解法三：其實(shí)這種解法我們也嘗試了，隊(duì)友說NTT搞了那么久，說明暴力應(yīng)該可以，不過最后只剩幾分鐘來不及找到合適的循環(huán)次數(shù)，http://www.cnblogs.com/cshg/p/5905398.html

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轉(zhuǎn)載于:https://www.cnblogs.com/youmi/p/5905679.html

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