題目

請計算C[k]=sigma(a[i]*b[i-k]) 其中 k < = i < n ，並且有 n < = 10 ^ 5。 a,b中的元素均為小於等於100的非負整數。

輸入格式

第一行一個整數N,接下來N行，第i+2..i+N-1行，每行兩個數，依次表示a[i],b[i] (0 < = i < N)。

輸出格式

輸出N行，每行一個整數，第i行輸出C[i-1]。

輸入樣例

5

3 1

2 4

1 1

2 4

1 4

輸出樣例

24

12

10

6

1

題解

和2179幾乎一模一樣
由於卷積的定義要求下標之和為常數，我們嘗試將原式變形，發現只要將a或者b反過來存就可以了

#include<iostream>
 

#include<cmath>
#include<cstdio>
#include<cstring>
#include<complex>
#include<algorithm>
#define pi acos(-1)
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<‘ ‘; puts("");
 

using namespace std;
const int maxn = 400005,maxm = 100005,INF = 1000000000;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57) {if (c == ‘-‘) flag = -1; c = getchar();}
    while (c >= 48 && c <= 57) {out = (out << 3) + (out << 1) + c - ‘0‘; c = getchar();}
    return
 
 out * flag;
}
typedef complex<double> E;
E a[maxn],b[maxn];
int n,m,L,R[maxn];
void fft(E* a,int f){
    for (int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]);
    for (int i = 1; i < n; i <<= 1){
        E wn(cos(pi / i),f * sin(pi / i));
        for (int j = 0; j < n; j += (i << 1)){
            E w(1,0);
            for (int k = 0; k < i; k++,w *= wn){
                E x = a[j + k],y = w * a[j + k + i];
                a[j + k] = x + y; a[j + k + i] = x - y;
            }
        }
    }
    if (f == -1) for (int i = 0; i < n; i++) a[i] /= n;
}
int main(){
    n = read(); n--;
    for (int i = 0; i <= n; i++){a[n - i] = read();b[i] = read();}
    m = n << 1; for (n = 1; n <= m; n <<= 1) L++;
    for (int i = 0; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
    fft(a,1); fft(b,1);
    for (int i = 0; i <= n; i++) a[i] *= b[i];
    fft(a,-1);
    for (int i = (m >> 1); i >= 0; i--) printf("%d\n",(int)(a[i].real() + 0.1));
    return 0;
}

BZOJ2194 快速傅立葉之二 【fft】