bzoj2194: 快速傅立叶之二
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 using namespace std; 7 const int maxn=400005; 8 const double PI=acos(-1); 9 struct node{ 10 double real,imag; 11 void clear(){real=imag=0;} 12 node operator +(const node &x){return (node){real+x.real,imag+x.imag};} 13 node operator -(const node &x){return (node){real-x.real,imag-x.imag};} 14 node operator *(const node &x){return (node){real*x.real-imag*x.imag,real*x.imag+imag*x.real};} 15 }a[maxn],b[maxn],c[maxn],t1,t2,w,wn; 16 int m,n,len,rev[maxn],ans[maxn]; 17 void read(int &x){ 18 x=0; int f=1; char ch; 19 for (ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') f=-1; 20 for (;isdigit(ch);ch=getchar()) x=x*10+ch-'0'; x*=f; 21 } 22 int Rev(int x){ 23 int temp=0; 24 for (int i=1;i<=len;i++) temp<<=1,temp+=(x&1),x>>=1; 25 return temp; 26 } 27 void FFT(node *a,int op){ 28 for (int i=0;i<n;i++) if (i<rev[i]) swap(a[i],a[rev[i]]); 29 for (int s=2;s<=n;s<<=1){ 30 wn=(node){cos(2*op*PI/s),sin(2*op*PI/s)}; 31 for (int i=0;i<n;i+=s){ 32 w=(node){1,0}; 33 for (int j=i;j<i+s/2;j++,w=w*wn){ 34 t1=a[j],t2=w*a[j+s/2]; 35 a[j]=t1+t2,a[j+s/2]=t1-t2; 36 } 37 } 38 } 39 } 40 int main(){ 41 read(m); n=1,len=0; 42 while (n<(m<<1)) n<<=1,len++; 43 for (int i=0;i<n;i++) rev[i]=Rev(i); 44 for (int x,i=0;i<m;i++) read(x),a[i].real=x,read(x),b[m-1-i].real=x; 45 FFT(a,1),FFT(b,1); 46 for (int i=0;i<n;i++) c[i]=a[i]*b[i]; 47 FFT(c,-1); 48 for (int i=0;i<n;i++) ans[i]=(int)round(c[i].real/n); 49 for (int i=m-1;i<2*m-1;i++) printf("%d\n",ans[i]); 50 return 0; 51 }
题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=2194
题目大意：请计算C[k]=sigma(a[i]*b[i-k]) 其中 k < = i < n ，并且有 n < = 10 ^ 5。 a,b中的元素均为小于等于100的非负整数。
做法：考虑把b数组翻转，Ck的计算就成为了裸的卷积，对于这种题目，翻转是个重要的手段。
