Time Limit: 16000/8000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 999 Accepted Submission(s): 434
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
Source
2016 Multi-University Training Contest 1
Recommend
wange2014
一段长为x的项链作为一个整体，有a[x]种装饰方案。可以把不同的整体连接起来。问长为n的项链共有多少种方案。
动态规划 分治FFT
设f[i]为长为i的方案数，很明显 $ f[i]=\sum_{j=1}^{i} a[j]*f[i-j] $
模数是313，这数的原根是啥啊？不知道。模数这么小，用FFT就可以了。
PS1 注意读入a[]的时候就要顺手取模，不然很容易乘爆炸
PS2 我也不知道为什么我要多输出一个换行符，白WA了三次才看到
1 #include<iostream>
2 #include<cstdio>
3 #include<algorithm>
4 #include<cstring>
5 #include<cmath>
6 #define LL long long
7 using namespace std;
8 const double pi=acos(-1.0);
9 const int mod=313;
10 const int mxn=200010;
11 int read(){
12 int x=0,f=1;char ch=getchar();
13 while(ch<‘0‘ || ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
14 while(ch>=‘0‘ && ch<=‘9‘){x=x*10-‘0‘+ch;ch=getchar();}
15 return x*f;
16 }
17 struct com{
18 double x,y;
19 com operator + (const com &b){return (com){x+b.x,y+b.y};}
20 com operator - (const com &b){return (com){x-b.x,y-b.y};}
21 com operator * (const com &b){return (com){x*b.x-y*b.y,x*b.y+y*b.x};}
22 com operator / (const double v){return (com){x/v,y/v};}
23 }a[mxn<<2],b[mxn<<2];
24 int N,len,rev[mxn<<2];
25 void FFT(com *a,int flag){
26 for(int i=0;i<N;i++)if(i<rev[i])swap(a[i],a[rev[i]]);
27 for(int i=1;i<N;i<<=1){
28 com wn=(com){cos(pi/i),flag*sin(pi/i)};
29 int p=i<<1;
30 for(int j=0;j<N;j+=p){
31 com w=(com){1,0};
32 for(int k=0;k<i;k++,w=w*wn){
33 com x=a[j+k],y=w*a[j+k+i];
34 a[j+k]=x+y;
35 a[j+k+i]=x-y;
36 }
37 }
38 }
39 if(flag==-1)
40 for(int i=0;i<N;i++)a[i].x/=N;
41 return;
42 }
43 int n,w[mxn];
44 LL f[mxn];
45 void solve(int l,int r){
46 if(l==r){(f[l]+=w[l])%=mod;return;}
47 int mid=(l+r)>>1;
48 solve(l,mid);
49 int i,j,m=(r-l+1);
50 for(N=1,len=0;N<=m;N<<=1)++len;
51 for(i=0;i<N;i++)rev[i]=(rev[i>>1]>>1)|((i&1)<<(len-1));
52 //
53 for(i=l;i<=mid;i++){a[i-l]=(com){f[i],0};}
54 for(i=mid-l+1;i<N;i++)a[i]=(com){0,0};
55 for(i=0;i<N;i++)b[i]=(com){w[i+1],0};
56 //
57 FFT(a,1);FFT(b,1);
58 for(i=0;i<N;i++)a[i]=a[i]*b[i];
59 FFT(a,-1);
60 for(i=mid+1;i<=r;i++){
61 (f[i]+=((LL)(a[i-l-1].x+0.5))%mod)%=mod;
62 }
63 solve(mid+1,r);
64 return;
65 }
66 int main(){
67 int i,j;
68 while(scanf("%d",&n)!=EOF && n){
69 memset(f,0,sizeof f);
70 for(i=1;i<=n;i++)w[i]=read()%mod;//
71 solve(1,n);
72 printf("%lld\n",f[n]%mod);
73 }
74 return 0;
75 }