
dp[n] = sigma(a[i]*dp[n-i]), 给出a1.....an, 求dp[n]。 n为1e5.
这个式子的形式显然是一个卷积, 所以可以用fft来优化一下, 但是这样也是会超时的。 所以可以用cdq分治来优化。
cdq分治就是处理(l, mid)的时候, 将dp[l]...dp[mid]对dp[mid+1]...dp[r]做的贡献都算出来。
#include <bits/stdc++.h> using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<11
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-;
const int mod = 1e9+;
const int inf = ;
const int dir[][] = { {-, }, {, }, {, -}, {, } };
typedef complex <double> cmx;
const int maxn = 2e5;
cmx x[maxn], y[maxn];
int a[maxn], dp[maxn], n;
void change(cmx x[], int len) {
int i, j, k;
for(i = , j = len/; i < len - ; i++) {
if(i < j)
swap(x[i], x[j]);
k = len / ;
while(j >= k) {
j -= k;
k /= ;
}
if(j < k)
j += k;
}
}
void fft(cmx x[], int len, int on) {
change(x, len);
for(int i = ; i <= len; i <<= ) {
cmx wn(cos(-on * * PI/i), sin(-on * * PI/i));
for(int j = ; j < len; j += i) {
cmx w(, );
for(int k = j; k < j + i/; k++) {
cmx u = x[k];
cmx v = x[k + i/]*w;
x[k] = u + v;
x[k+i/] = u - v;
w *= wn;
}
}
}
if(on == -) {
for(int i = ; i < len; i++)
x[i] /= len;
}
}
void cdq(int l, int r)
{
if(l == r) {
dp[l] += a[l];
dp[l] %= mod;
return ;
}
int mid = l+r>>;
cdq(l, mid);
int len = ;
while(len <= (r-l+))
len <<= ;
for(int i = ; i < len; i++) {
x[i] = y[i] = cmx(, );
}
for(int i = l; i <= mid; i++) {
x[i-l] = cmx(dp[i], );
}
for(int i = ; i <= r-l; i++) {
y[i-] = cmx(a[i], );
}
fft(x, len, );
fft(y, len, );
for(int i = ; i < len; i++)
x[i] *= y[i];
fft(x, len, -);
for(int i = mid+; i <= r; i++) {
dp[i] += (int)(x[i-l-].real()+0.5);
dp[i] %= mod;
}
cdq(mid+, r);
}
int main()
{
while(scanf("%d", &n) && n) {
mem(dp);
for(int i = ; i <= n; i++) {
scanf("%d", &a[i]);
a[i] %= mod;
}
cdq(, n);
cout<<dp[n]<<endl;
}
}