HDU5730 Shell Necklace

標籤：cstring   span   diff   fft   algorithm   nbsp   stream   each   ==   

Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 999    Accepted Submission(s): 434

Problem DescriptionPerhaps 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. 

 

InputThere 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.
 

 

OutputFor 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 Input31 3 742 2 2 2 0 

 

Sample Output1454 HintFor 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. 

 

AuthorHIT 

 

Source2016 Multi-University Training Contest 1 

 

Recommendwange2014

 

一段長為x的項鏈作為一個整體，有a[x]種裝飾方案。可以把不同的整體串連起來。問長為n的項鏈共有多少種方案。

 

動態規劃 分治FFT

設f[i]為長為i的方案數，很明顯 $ f[i]=\sum_{j=1}^{i} a[j]*f[i-j] $

模數是313，這數的原根是啥啊？不知道。模數這麼小，用FFT就可以了。

 

PS1 注意讀入a[]的時候就要順手模數，不然很容易乘爆炸

PS2 我也不知道為什麼我要多輸出一個分行符號，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 }

 

HDU5730 Shell Necklace