bzoj 2194 [快速傅里叶变换]

题目

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。

输入

第一行一个整数N,接下来N行，第i+2..i+N-1行，每行两个数，依次表示a[i],b[i] (0 < = i < N)。

输出

输出N行，每行一个整数，第i行输出C[i-1]。

样例输入

5
3 1
2 4
1 1
2 4
1 4

样例输出

24
12
10
6
1

分析

分析题目,将题中给出的式子简单变换,将b数组反转:

Ck=∑i=kn(ai⋅bi−k)→Ck=∑i=kn(ai⋅bn−i−k)

将c数组反转:

Cn−k=∑i=kn(ai⋅bk−i)

即转化成了卷积的形式,直接FFT求解即可

完整代码

#include<bits/stdc++.h> #define pi acos(-1.0)#define maxn 300010//#define DEBUGusing namespace std;int n;complex<double> a[maxn], b[maxn];inline int read(){    char ch;    int read = 0, sign = 1;    do        ch = getchar();    while ((ch<'0'||ch>'9')&&ch!='-');    if (ch == '-') sign = -1, ch = getchar();    while (ch >= '0' && ch <= '9')    {        read = read * 10 + ch - '0';        ch = getchar();    }    return read*sign;}inline int Power2(int x){    int x0;    for (x0 = 1; x0 < x; x0 <<= 1);    return x0;}inline int lg(int n){    int l = 0;    if (n == 0) return l;    for (int x0 = 1; x0 <= n; x0 <<= 1) l++;    return l;}inline int rev(int x, int n){    int out = 0;    while (n--) out = (out + (x & 1)) << 1, x >>= 1;    return out>>1;}void FFT(complex<double> a[],int n, int flag){    complex<double> A[n+1];    for (int i = 0, l = lg(n - 1); i < n; ++i) A[rev(i, l)] = a[i];#ifdef DEBUG    int l=lg(n-1);    cerr<<"l="<<l<<endl;    for(int i=0;i<n;++i) cerr<<rev(i,l)<<" ";    cerr<<endl;#endif     for (int i = 2; i <= n; i <<= 1)    {        complex<double> dw(cos(2*pi/i),sin(flag*2*pi/i));        for (int j = 0; j < n; j += i)        {            complex<double> w(1.0, 0);            for (int k = 0; k < (i >> 1); k++, w *= dw)            {                complex<double> u = A[j + k];                complex<double> t = w*A[j + k + (i >> 1)];                A[j + k] = u + t;                A[j + k + (i >> 1)] = u - t;            }        }         if (flag == -1)            for (int i = 0; i < n; i++) a[i].real() = int(A[i].real() / n + 0.5);        else            for (int i = 0; i < n; i++) a[i] = A[i];    }}int main(){    n = read();    for (int i = 0; i < n; ++i)        a[n-i-1] = read(), b[i] = read();#ifdef DEBUG    for(int i=0;i<n;++i) cerr<<a[i].real()<<" ";    cerr<<endl;    for(int i=0;i<n;++i) cerr<<b[i].real()<<" ";     cerr<<endl;#endif     int length = Power2(n);#ifdef DEBUG    cerr<<"length="<<length<<endl;#endif    FFT(a, length, 1);    FFT(b, length, 1);    for (int i = 0; i < length; ++i) a[i] *= b[i];    FFT(a, length, -1);    for (int i = n-1; i >= 0; --i) cout << a[i].real() << endl;    return 0;}