HDU 5730 - Shell Necklace

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 =  * atan(1.0);
 const int MAXN = ;
 const int MOD = ;
 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 = , j = len/; i < len-; i++)
     {
         if (i < j) swap(y[i], y[j]);
         k = len / ;
         while (j >= k)
         {
             j -= k;
             k /= ;
         }
         if (j < k) j += k;
     }
 }
 void FFT(Complex y[], int len,int on)
 {
     Change(y, len);
     for (int h = ; h <= len; h <<= )
     {
         Complex wn( cos(-on**PI/h), sin(-on**PI/h) );
         for (int j = ; j < len; j +=h)
         {
             Complex w(, );
             for (int k = j; k < j + h/; k++)
             {
                 Complex u = y[k];
                 Complex t = w * y[k + h/];
                 y[k] = u + t;
                 y[k + h/] = u - t;
                 w = w * wn;
             }
         }
     }
     if (on == -)
         for (int i = ; i < len; i++)
             y[i].x /= len;
 }
 int t, n;
 Complex x[MAXN], y[MAXN];
 int a[MAXN/], dp[MAXN/];
 void CDQ(int l, int r)
 {
     if (l == r) { dp[l] = (dp[l] + a[l]) % MOD; return; }
     int mid = (l + r) >> ;
     CDQ(l, mid);//处理前半段
     int len = , len1 = mid - l + , len2 = r - l + ;
     while(len < len2) len <<= ;
     for (int i = ; i < len1; i++) x[i] = Complex(dp[i + l], );
     for (int i = len1; i < len; i++) x[i] = Complex(, );
     for (int i = ; i < len2; i++) y[i] = Complex(a[i], );
     for (int i = len2; i < len; i++) y[i] = Complex(, );
     FFT(x, len, );
     FFT(y, len, );
     for (int i = ; i < len; i++) x[i] = x[i] *y[i];
     FFT(x, len, -);
     for (int i = mid+; i <= r; i++)//更新贡献
     {
         dp[i] = (int)(dp[i] + x[i - l].x + 0.5) %MOD;
     }
     CDQ(mid + , r);//处理后半段
 }
 int main()
 {
     while(~scanf("%d",&n) && n)
     {
         for (int i = ; i <= n; i++)
         {
             scanf("%d",&a[i]);
             a[i] %= MOD;
             dp[i] = ;
         }
         CDQ(, n);
         printf("%d\n", dp[n]);
     }
 }