hihocoder #1388 : Periodic Signal NTT&FFT
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传送门:hihocoder #1388 : Periodic Signal
先来几个大牛传送门: (模板) NTT long long 版
解法一:因为我们知道FFT会精度不够,所以坚持用NTT,但是模数不够大,然后就一直GG,看来我们的搜索姿势也有问题,居然没有搜到上面大神的板子,真的是GG
http://www.cnblogs.com/WABoss/p/5903927.html
/************************************************************** Problem: User: youmi Language: C++ Result: Accepted Time: 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()///卷积,点乘,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_JUDGE freopen("in.txt","r",stdin); #endif int 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; } }
解法二:这个解法确实很漂亮,比赛的时候一直徘徊找一个更大的 模数,然后就GG了,http://www.cnblogs.com/smartweed/p/5903838.html
解法三:其实这种解法我们也尝试了,队友说NTT搞了那么久,说明暴力应该可以,不过最后只剩几分钟来不及找到合适的循环次数,http://www.cnblogs.com/cshg/p/5905398.html
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