[BZOJ2194] 快速傅立叶之二

2194: 快速傅立叶之二

Time Limit: 10 Sec  Memory Limit: 259 MB
Submit: 1993  Solved: 1201
[Submit][Status][Discuss]

Description

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。

 

Input

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)。

Output

输出N行，每行一个整数，第i行输出C[i-1]。

Sample Input

5
3 1
2 4
1 1
2 4
1 4

Sample Output

24
12
10
6
1

 

先把f逆序，卷积之后再逆序输出

#include <bits/stdc++.h>
using namespace std;
const int maxn = 262144;
struct comp{
    double x, y;
    comp(double _x = 0, double _y = 0){
        x = _x;
        y = _y;
    }
    friend comp operator * (const comp &a, const comp &b){
        return comp(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x);
    }
    friend comp operator + (const comp &a, const comp &b){
        return comp(a.x + b.x, a.y + b.y);
    }
    friend comp operator - (const comp &a, const comp &b){
        return comp(a.x - b.x, a.y - b.y);
    }
}f[maxn], g[maxn];
int rev[maxn];
void dft(comp A[], int len, int kind){
    for(int i = 0; i < len; i++){
        if(i < rev[i]){
            swap(A[i], A[rev[i]]);
        }
    }
    for(int i = 1; i < len; i <<= 1){
        comp wn(cos(acos(-1.0) / i), kind * sin(acos(-1.0) / i));
        for(int j = 0; j < len; j += (i << 1)){
            comp tmp(1, 0);
            for(int k = 0; k < i; k++){
                comp s = A[j + k], t = tmp * A[i + j + k];
                A[j + k] = s + t;
                A[i + j + k] = s - t;
                tmp = tmp * wn;
            }
        }
    }
    if(kind == -1) for(int i = 0; i < len; i++) A[i].x /= len;
}
int main(){
    int n, len, L = 0;
    scanf("%d", &n); 
    for(len = 1; len < n + n - 1; len <<= 1, L++);
    for(int i = 0; i < len; i++){
        rev[i] = rev[i >> 1] >> 1 | (i & 1) << L - 1;
    }
    for(int i = 0; i < n; i++)
        scanf("%lf%lf", &f[n - i - 1].x, &g[i].x);
    dft(f, len, 1); dft(g, len, 1);
    for(int i = 0; i < len; i++) f[i] = f[i] * g[i];
    dft(f, len, -1);
    for(int i = n - 1; ~i; i--) printf("%d\n", (int)(f[i].x + 0.5));
    return 0; 
}