bzoj2194 快速傅立叶之二
Posted wxyww
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problem
给出两个长度为n的数列a,b。求一个数列c满足:[c[k] = sumlimits_{i = k} ^ na[i]b[i - k]]
(nle 10^5)
solution
长得和卷积很像,观察一下卷积的形式:(c[k]=sumlimits_{i=0}^ia[i]b[k-i])
所以先把b数组翻转过来。
然后所求的式子就变成了(c[k]=sumlimits_{i=k}^na[i]b[n+k-i])
将(n+k)看做整体,那么形式就和卷积比较像了,令(d[n+k]=c[k]=sumlimits_{i=k}^na[i]b[n + k - i])。因为a的(n+1)到(n+k)位置均为0,b的(0)到(k-1)位置均为0.
所以上式也可以写为(d[n+k]=sumlimits_{i=0}^{n+k}a[i]b[n+k-i])。
这样就可以借助FFT计算了。
注意最后的答案是问c[k],也就是d[k+n]
code
/*
* @Author: wxyww
* @Date: 2020-01-22 08:04:15
* @Last Modified time: 2020-01-22 08:24:04
*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
#include<ctime>
using namespace std;
typedef long long ll;
const int N = 400010;
const double pi = 3.1415926535897931;
ll read() {
ll x=0,f=1;char c=getchar();
while(c<'0'||c>'9') {
if(c=='-') f=-1;
c=getchar();
}
while(c>='0'&&c<='9') {
x=x*10+c-'0';
c=getchar();
}
return x*f;
}
struct complex {
double x,y;
}A[N],B[N];
complex operator * (const complex &A,const complex &B) {
return (complex){A.x * B.x - A.y * B.y,A.x * B.y + A.y * B.x};
}
complex operator + (const complex &A,const complex &B) {
return (complex){A.x + B.x,A.y + B.y};
}
complex operator - (const complex &A,const complex &B) {
return (complex){A.x - B.x,A.y - B.y};
}
int rev[N];
void FFT(complex *a,int n,int xs) {
for(int i = 0;i <= n;++i) if(rev[i] > i) swap(a[rev[i]],a[i]);
for(int m = 2;m <= n;m <<= 1) {
complex w1 = (complex){cos(2 * pi / m),xs * sin(2 * pi / m)};
for(int i = 0;i < n;i += m) {
complex w = (complex){1,0};
for(int k = 0;k < (m >> 1);++k) {
complex u = a[i + k];
complex t = w * a[i + k + (m >> 1)];
a[i + k] = u + t;
a[i + k + (m >> 1)] = u - t;
w = w * w1;
}
}
}
}
int main() {
int n = read();
for(int i = 0;i < n;++i) {
A[i].x = read();B[n - i - 1].x = read();
}
--n;
int tot = 1;
while(tot <= (n << 1)) tot <<= 1;
for(int i = 0;i <= tot;++i) rev[i] = (rev[i >> 1] >> 1) | (i & 1 ? tot >> 1 : 0);
FFT(A,tot,1);FFT(B,tot,1);
for(int i = 0;i <= tot;++i) A[i] = A[i] * B[i];
FFT(A,tot,-1);
for(int i = 0;i <= n;++i) printf("%d
",(int)round(A[i + n].x / tot));
return 0;
}
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