HDU 5730

题意：

　　给出连续的1-n个珠子的涂色方法 a[i](1<=i<=n)， 问长度为n的珠链共有多少种涂色方案

分析：

　　可以得到DP方程： DP[n] = ∑(i=1,n) (DP[n-i]*a[i]).

　　该方程为卷积形式，故 CDQ + FFT

　　

　　CDQ： 将 [l,r] 二分， 先得到[l,mid]的答案，再更新[l,mid]对[mid+1,r]的贡献.

　　　　　  对任意 DP[j](mid+1 <= j <= r), [l,mid] 对其贡献为 ∑(i=l,mid) (DP[i]*a[j - i]) , 即多项式 DP 与 a 相乘后次数为j项.

　　FFT： 优化多项式相乘.

（1 和 l 看不清的也就这破博客园了，代码还是粘下来的好，= =）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const double PI = 4 * atan(1.0);
const int MAXN = 200005;
const int MOD = 313;
struct Complex
{
    double x, y;
    Complex(double xx = 0.0, double yy = 0.0) : x(xx), y(yy) {}
    Complex operator - (const Complex &b) const 
    {
        return Complex(x - b.x, y - b.y);
    }
    Complex operator + (const Complex &b) const
    {
        return Complex(x + b.x, y + b.y);
    }
    Complex operator * (const Complex &b) const
    {
        return Complex(x*b.x - y*b.y, x*b.y + y*b.x);
    }
};
void Change(Complex y[], int len)
{
    int i, j, k;
    for (i = 1, j = len/2; i < len-1; i++)
    {
        if (i < j) swap(y[i], y[j]);
        k = len / 2;
        while (j >= k)
        {
            j -= k;
            k /= 2;
        }
        if (j < k) j += k;
    }
}
void FFT(Complex y[], int len,int on)
{
    Change(y, len);
    for (int h = 2; h <= len; h <<= 1)
    {
        Complex wn( cos(-on*2*PI/h), sin(-on*2*PI/h) );
        for (int j = 0; j < len; j +=h)
        {
            Complex w(1, 0);
            for (int k = j; k < j + h/2; k++)
            {
                Complex u = y[k];
                Complex t = w * y[k + h/2];
                y[k] = u + t;
                y[k + h/2] = u - t;
                w = w * wn;                
            }
        }
    } 
    if (on == -1)
        for (int i = 0; i < len; i++)
            y[i].x /= len;
}
int t, n;
Complex x[MAXN], y[MAXN];
int a[MAXN/2], dp[MAXN/2];
void CDQ(int l, int r)
{
    if (l == r) { dp[l] = (dp[l] + a[l]) % MOD; return; } 
    int mid = (l + r) >> 1; 
    CDQ(l, mid);//处理前半段 
    int len = 1, len1 = mid - l + 1, len2 = r - l + 1;
    while(len < len2) len <<= 1;
    for (int i = 0; i < len1; i++) x[i] = Complex(dp[i + l], 0);
    for (int i = len1; i < len; i++) x[i] = Complex(0, 0);
    for (int i = 0; i < len2; i++) y[i] = Complex(a[i], 0);
    for (int i = len2; i < len; i++) y[i] = Complex(0, 0); 
    FFT(x, len, 1);
    FFT(y, len, 1);
    for (int i = 0; i < len; i++) x[i] = x[i] *y[i];
    FFT(x, len, -1);
    for (int i = mid+1; i <= r; i++)//更新贡献 
    {
        dp[i] = (int)(dp[i] + x[i - l].x + 0.5) %MOD;
    }
    CDQ(mid + 1, r);//处理后半段 
}
int main()
{
    while(~scanf("%d",&n) && n)
    {
        for (int i = 1; i <= n; i++)
        {
            scanf("%d",&a[i]);
            a[i] %= MOD;
            dp[i] = 0;
        }
        CDQ(1, n);
        printf("%d
", dp[n]);
    }
}