HDU5730 Shell Necklace

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

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define LL long long
using namespace std;
const double pi=acos(-1.0);
const int mod=313;
const int mxn=200010;
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0' && ch<='9'){x=x*10-'0'+ch;ch=getchar();}
    return x*f;
}
struct com{
    double x,y;
    com operator + (const com &b){return (com){x+b.x,y+b.y};}
    com operator - (const com &b){return (com){x-b.x,y-b.y};}
    com operator * (const com &b){return (com){x*b.x-y*b.y,x*b.y+y*b.x};}
    com operator / (const double v){return (com){x/v,y/v};}
}a[mxn<<2],b[mxn<<2];
int N,len,rev[mxn<<2];
void FFT(com *a,int flag){
    for(int i=0;i<N;i++)if(i<rev[i])swap(a[i],a[rev[i]]);
    for(int i=1;i<N;i<<=1){
        com wn=(com){cos(pi/i),flag*sin(pi/i)};
        int p=i<<1;
        for(int j=0;j<N;j+=p){
            com w=(com){1,0};
            for(int k=0;k<i;k++,w=w*wn){
                com x=a[j+k],y=w*a[j+k+i];
                a[j+k]=x+y;
                a[j+k+i]=x-y;
            }
        }
    }
    if(flag==-1)
        for(int i=0;i<N;i++)a[i].x/=N;
    return;
}
int n,w[mxn];
LL f[mxn];
void solve(int l,int r){
    if(l==r){(f[l]+=w[l])%=mod;return;}
    int mid=(l+r)>>1;
    solve(l,mid);
    int i,j,m=(r-l+1);
    for(N=1,len=0;N<=m;N<<=1)++len;
    for(i=0;i<N;i++)rev[i]=(rev[i>>1]>>1)|((i&1)<<(len-1));
    //
    for(i=l;i<=mid;i++){a[i-l]=(com){f[i],0};}
    for(i=mid-l+1;i<N;i++)a[i]=(com){0,0};
    for(i=0;i<N;i++)b[i]=(com){w[i+1],0};
    //
    FFT(a,1);FFT(b,1);
    for(i=0;i<N;i++)a[i]=a[i]*b[i];
    FFT(a,-1);
    for(i=mid+1;i<=r;i++){
        (f[i]+=((LL)(a[i-l-1].x+0.5))%mod)%=mod;
    }
    solve(mid+1,r);
    return;
}
int main(){
    int i,j;
    while(scanf("%d",&n)!=EOF && n){
        memset(f,0,sizeof f);
        for(i=1;i<=n;i++)w[i]=read()%mod;//
        solve(1,n);
        printf("%lld
",f[n]%mod);
    }
    return 0;
}