Shell Necklace

Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 866    Accepted Submission(s): 383 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 n non-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

题意：

一段长为i的项链有a[i]中装饰方法，问长度为n的项链有多少种装饰方式。

思路：

容易推出，dp[i]=∑dp[j]*a[i-j],(1<=j<=i-1)那么这样就刚好符合卷积的运算，这样就可以愉快地使用fft了，不过数量级在1e5，所以应该采用分治来处理，算法复杂度nlognlogn。

#include<iostream> #include<cstdio> #include<cstdlib> #include<algorithm> #include<cstring> #include <cmath> using namespace std; const double PI = acos(-1.0); const int maxn = 100010 * 4; struct complex { double r,i; complex(double _r = 0.0,double _i = 0.0) { r = _r; i = _i; } complex operator +(const complex &b) { return complex(r+b.r,i+b.i); } complex operator -(const complex &b) { return complex(r-b.r,i-b.i); } complex operator *(const complex &b) { return complex(r*b.r-i*b.i,r*b.i+i*b.r); } }; /* * 进行FFT和IFFT前的反转变换。 * 位置i和 （i二进制反转后位置）互换 * len必须去2的幂 */ 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]); //交换互为小标反转的元素，i<j保证交换一次 //i做正常的+1，j左反转类型的+1,始终保持i和j是反转的 k = len/2; while( j >= k) { j -= k; k /= 2; } if(j < k) j += k; } } /* * 做FFT * len必须为2^k形式， * on==1时是DFT，on==-1时是IDFT */ 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].r /= len; } complex x1[maxn], x2[maxn]; void getfft(int arg1[], int len1, int arg2[], int len2, int fft_out[])//模板：第一个数组，数组元素数量。 第二个数组，数组元素数量，输出数组 { int len = 1; while(len < len1*2 || len < len2*2)len<<=1; for(int i = 0;i < len1;i++) x1[i] = complex(arg1[i] , 0); for(int i = len1;i < len;i++) x1[i] = complex(0 , 0); for(int i = 0;i < len2;i++) x2[i] = complex(arg2[i] , 0); for(int i = len2;i < len;i++) x2[i] = complex(0, 0); fft(x1,len,1); fft(x2,len,1); for(int i = 0;i < len;i++) x1[i] = x1[i]*x2[i]; fft(x1,len,-1); for(int i = 0;i < len;i++) fft_out[i] = x1[i].r+0.5; } int n; const int mod = 313; int a[maxn], f[maxn], sum[maxn]; void init() { memset(f, 0, sizeof(f)); f[0]=1; for (int i = 1; i <= n; ++i) { scanf("%d", &a[i]); a[i] %= 313; } } void solve(int l ,int r) { if (l == r) { f[l] = (f[l] + a[l]) % mod; //f[l]一定已经求出了 return; } int mid = l + (r-l)/2; solve(l, mid); int len1 = mid -l + 1; int len2 = r - l; getfft(f+l, len1, a+1, len2, sum); for (int i = mid + 1; i <= r; ++ i) f[i] = (f[i] + sum[i-(l+1)]) % mod; solve(mid + 1, r); } void doit() { solve(1, n); printf("%d\n", f[n]); } int main() { while (~scanf("%d", &n)) { if (!n) break; init(); doit(); } return 0; }