2016 Multi-University Training Contest 1 H.Shell Necklace

Shell Necklace

Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 534    Accepted Submission(s): 227

Problem Description

Perhaps the sea‘s definition of a shell is the pearl. However, in my view, a shell necklace with n beautiful shells contains the most sincere feeling for my best lover Arrietty, but even that is not enough.

Suppose the shell necklace is a sequence of shells (not a chain end to end). Considering i continuous shells in the shell necklace, I know that there exist different schemes to decorate the i shells together with one declaration of love.

I want to decorate all the shells with some declarations of love and decorate each shell just one time. As a problem, I want to know the total number of schemes.

 

Input

There are multiple test cases(no more than 20 cases and no more than 1 in extreme case), ended by 0.

For each test cases, the first line contains an integer n, meaning the number of shells in this shell necklace, where 1≤n≤105. Following line is a sequence with nnon-negative integer a1,a2,…,an, and ai≤107 meaning the number of schemes to decorate i continuous shells together with a declaration of love.

 

Output

For each test case, print one line containing the total number of schemes module 313(Three hundred and thirteen implies the march 13th, a special and purposeful day).

 

Sample Input

3
1 3 7
4
2 2 2 2
0

 

Sample Output

14
54

Hint

For the first test case in Sample Input, the Figure 1 provides all schemes about it. The total number of schemes is 1 + 3 + 3 + 7 = 14.

 

Author

HIT

 

题意：
    一串由n颗珍珠组成的项链，连续的i个珍珠有ci种染色方式。
    问n颗珍珠有多少种染色方式？

题解：
    显然可知dp方程
    f[i] = sigma(f[i - j] * a[j])
    由于n很大，可以考虑分治fft解决。

    说说这个分治fft：
    如果我们需要计算出l～r的f的值。
    假设我们已经计算出（l，mid）的f的值
    观察方程
    f[i] = sigma(f[j] * a[i - j]) 1 <= j < i
        = sigma(f[j] * a[i - j]) + sigma(f[k] * a[i - k])   1<= j <= mid, mid < k < i
    所以可以分开计算f[l~mid]对f[mid+1~r]的影响。
    具体过程可以看代码。
    使用时记得调整一下下标。

　　

 class Complex {
     public :
         double real, image;

         Complex(double real = ., double image = .):real(real), image(image) {}
         Complex(const Complex &t):real(t.real), image(t.image) {}

         Complex operator +(const Complex &t) const {
             return Complex(real + t.real, image + t.image);
         }

         Complex operator -(const Complex &t) const {
             return Complex(real - t.real, image - t.image);
         }

         Complex operator *(const Complex &t) const {
             return Complex(real * t.real - image * t.image,
                         real * t.image + t.real * image);
         }
 };

 const int N = , MOD = ;
 const double PI = acos(-.);

 class FFT {
     /**
      * 1. Need define PI
      * 2. Need define class Complex
      * 3. tmp is need for fft, so define a N suffice it
      * 4. dig[30] -> (1 << 30) must bigger than N
      * */
     private :
         static Complex tmp[N];
         static int revNum[N], dig[];

     public :
         static void init(int n) {
             int len = ;
             for(int t = n - ; t; t >>= ) ++len;
             for(int i = ; i < n; i++) {
                 revNum[i] = ;
                 for(int j = ; j < len; j++) dig[j] = ;
                 for(int idx = , t = i; t; t >>= ) dig[idx++] = t & ;
                 for(int j = ; j < len; j++)
                     revNum[i] = (revNum[i] << ) | dig[j];
             }
         }

         static int rev(int x) {
             return revNum[x];
         }

         static void fft(Complex a[], int n, int flag) {
             /**
              * flag = 1 -> DFT
              * flag = -1 -> IDFT
              * */
             for(int i = ; i < n; ++i) tmp[i] = a[rev(i)];
             for(int i = ; i < n; ++i) a[i] = tmp[i];
             for(int i = ; i <= n; i <<= ) {
                 Complex wn(cos( * PI / i), flag * sin( * PI / i));
                for(int k = , half = i / ; k < n; k += i) {
                    Complex w(., .);
                    for(int j = k; j < k + half; ++j) {
                        Complex x = a[j], y = w * a[j + half];
                        a[j] = x + y, a[j + half] = x - y;
                        w = w * wn;
                    }
                }
             }
             if(flag == -) {
                 for(int i = ; i < n; ++i) a[i].real /= n;
             }
         }

         static void dft(Complex a[], int n) {
             fft(a, n, );
         }

         static void idft(Complex a[], int n) {
             fft(a, n, -);
         }
 };
 Complex FFT::tmp[N];
 int FFT::revNum[N], FFT::dig[];

 int n, arr[N];
 int f[N];
 Complex A[N], B[N];

 inline void divideAndConquer(int lef, int rig) {
     if(lef >= rig) return;
     int mid = (lef + rig) >> ;
     divideAndConquer(lef, mid);

     int m;
     for(m = ; m < (rig - lef + ) * ; m <<= );
     for(int i = ; i < m; ++i) A[i] = B[i] = Complex();
     for(int i = lef; i <= mid; ++i) A[i - lef] = Complex(f[i]);
     int len = min(n, m - );
     for(int i = ; i < len; ++i) B[i + ] = Complex(arr[i]);
     FFT::init(m);
     FFT::dft(A, m), FFT::dft(B, m);
     for(int i = ; i < m; ++i) A[i] = A[i] * B[i];
     FFT::idft(A, m);

     for(int i = mid + ; i <= rig; ++i)
         f[i] = (f[i] + ((ll)(A[i - lef].real + 0.5))) % MOD;

     divideAndConquer(mid + , rig);
 }

 inline void solve() {
     for(int i = ; i < n; ++i) arr[i] %= MOD;
     for(int i = ; i <= n; ++i) f[i] = ;
     f[] = ;
     divideAndConquer(, n);

     printf("%d\n", f[n]);
 }

 int main() {
     while(scanf("%d", &n) ==  && n) {
         for(int i = ; i < n; ++i) scanf("%d", &arr[i]);
         solve();
     }
     return ;
 }