【HDU5730 2016 Multi-University Training Contest 1H】【FFT + cdq 分治】 Shell Necklace f[i]=∑f[i-j] x a[j]

Shell Necklace

Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 290    Accepted Submission(s): 116

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

31 3 742 2 2 2 0

 

Sample Output

1454
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

 

Source

2016 Multi-University Training Contest 1

#include<stdio.h>#include<iostream>#include<string.h>#include<string>#include<ctype.h>#include<math.h>#include<set>#include<map>#include<vector>#include<queue>#include<bitset>#include<algorithm>#include<time.h>using namespace std;void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }#define MS(x,y) memset(x,y,sizeof(x))#define MC(x,y) memcpy(x,y,sizeof(x))#define MP(x,y) make_pair(x,y)#define ls o<<1#define rs o<<1|1typedef long long LL;typedef unsigned long long UL;typedef unsigned int UI;typedef double ld;template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }const int N = 1 << 17, Z = 313, ms63 = 0x3f3f3f3f;const double PI = acos(-1.0);struct Complex{double r, i;Complex(double r = 0, double i = 0) : r(r), i(i) {}Complex operator + (const Complex& t) const { return Complex(r + t.r, i + t.i); }Complex operator - (const Complex& t) const { return Complex(r - t.r, i - t.i); }Complex operator * (const Complex& t) const { return Complex(r * t.r - i * t.i, r * t.i + i * t.r); }}X[N], Y[N];void FFT(Complex y[], int n, int rev)//rev == 1时为DFT，rev == -1时为IDFT{for (int i = 1, j, k, t; i < n; ++i){for (j = 0, k = n >> 1, t = i; k; k >>= 1, t >>= 1) j = j << 1 | t & 1;if (i < j) swap(y[i], y[j]);}for (int s = 2, ds = 1; s <= n; ds = s, s <<= 1){Complex wn(cos(rev * 2 * PI / s), sin(rev * 2 * PI / s));for (int k = 0; k < n; k += s){Complex w(1, 0), t;for (int i = k; i < k + ds; ++i){y[i + ds] = y[i] - (t = w * y[i + ds]);y[i] = y[i] + t;w = w * wn;}}}if (rev == -1) for (int i = 0; i < n; ++i) y[i].r /= n;}int n, a[N], f[N];void cdq(int l, int r){if (l == r) { f[l] = (f[l] + a[l]) % Z; return; }int mid = l + r >> 1;cdq(l, mid);int len1 = mid - l + 1;int len2 = r - l + 1;int len = 1; while (len < len2) len <<= 1;//这里的数组大小应该是while(len < len1 + len2)for (int i = 0; i < len1; ++i) X[i] = Complex(f[l + i], 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) Y[i] = X[i] * Y[i];FFT(Y, len, -1);for (int i = mid + 1; i <= r; ++i)f[i] = f[i] + (int)(Y[i - l].r + 0.5) % Z;cdq(mid + 1, r);}int main(){while (~scanf("%d", &n), n){MS(f, 0);for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), a[i] %= Z;cdq(1, n);printf("%d\n", f[n]);}return 0;}/*【trick&&吐槽】在本题，我们做FFT多项式乘法的时候，X*Y的目标数组的长度要开多少呢？按照正常的乘法逻辑，我们要把数组开到len1+len2.然而，这道题我们只需要使用到len1【题意】有n（1e5）个数a[]，a[i]都是[1e7]范围的数。f[1]=0;f[i]=∑(f[i - j] * a[j]), j∈[1, i];让你求f[n] % Z【类型】fft + cdq分治【分析】这道题我们不能做n次fft，会重复计算，最终导致TLE。我们采取cdq分治的方式做加速。所谓cdq分治，利用了这道题目中的——想求f[i],需要加进所有f[j，j<i]对f[i]的贡献我们把区间分成了以下两块[l,mid] [mid+1,r]，我们先处理[l,mid]，然后把[l,mid]的影响向[mid+1,r]转移，再去处理[mid+1,r]。问题是，如何做转移呢？fft。回归fft如何做大数相乘运算v[0] v[1] v[2]u[0] u[1] u[2]我们乘出来之后变成了——(0*0) (0*1+1*0) (0*2+1*1*2*0) (1*2+2*1) (2*2)显然，有一侧为f[l],f[l+1],f[l+2],f[l+3],...,f[mid-1],f[mid]另一侧为a[0],a[1],a[2],...,a[mid-l]我们乘起来就变成了——(f[l] * a[0]) =不需要的数...(f[l] * a[mid+1-l] + f[l+1] * a[mid+1-l-1] + ...) = f[mid+1]...即，我们需要f[l]~f[mid]中的每个数同时需要a[0]~a[r-l]中的每个数然后FFT就可以得到结果【时间复杂度&&优化】尽管我们做了nlogn次FFT但是事实上相当于只做了logn层长度为n的FFT总的复杂度为O(nlognlogn)，可以无压力AC这道题的数组我们要开多大呢？100000的上log 1<<17即可*/