D. New Year and Ancient Prophecy

题目连接：

http://www.codeforces.com/contest/611/problem/C

Description

Limak is a little polar bear. In the snow he found a scroll with the ancient prophecy. Limak doesn't know any ancient languages and thus is unable to understand the prophecy. But he knows digits!

One fragment of the prophecy is a sequence of n digits. The first digit isn't zero. Limak thinks that it's a list of some special years. It's hard to see any commas or spaces, so maybe ancient people didn't use them. Now Limak wonders what years are listed there.

Limak assumes three things:

Years are listed in the strictly increasing order;

Every year is a positive integer number;

There are no leading zeros.

Limak is going to consider all possible ways to split a sequence into numbers (years), satisfying the conditions above. He will do it without any help. However, he asked you to tell him the number of ways to do so. Since this number may be very large, you are only asked to calculate it modulo 109 + 7.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 5000) — the number of digits.

The second line contains a string of digits and has length equal to n. It's guaranteed that the first digit is not '0'.

Output

Print the number of ways to correctly split the given sequence modulo 109 + 7.

Sample Input

123434

Sample Output

Hint

题意：

给你一个全是数字的字符串（长度5000），问你多少种划分方案，就可以使得这个字符串分割成了一个绝对递增序列。

题解

DP，dp[i][j]表示以i位置结尾，长度为j的字符串的方案数。转移很简单，就dp[i][j]+=dp[i-j]k，如果str[i-j+1][i]>str[i-j-j+1][i-j]的话，dp[i][j]+=dp[i-j][j]。

很显然，dp是n^3的，我们就可以用奇怪的手法去优化一下就好了，我是无脑后缀数组预处理优化的。

代码

#include<bits/stdc++.h>

using namespace std;

long long dp[5005][5005];

char str[5005];

const int mod = 1e9+7;

char s[5005];

struct Bit

{

    int lowbit(int x)

    {

        return x&(-x);

    }

    long long val[5005];

    int sz;

    void init(int sz){

        this->sz=sz;

        for(int i = 0 ; i <= sz ; ++ i) val[i] = 0 ;

    }

    void updata(int pos ,long long key)

    {

        while(pos<=sz){

            val[pos]+=key;

            if(val[pos]>=mod)

                val[pos]-=mod;

            pos+=lowbit(pos);

        }

    }

    long long query(int pos)

    {

        long long res=0;

        while(pos>0)

        {

            res+=val[pos];

            if(res>=mod)res-=mod;

            pos-=lowbit(pos);

        }

        return res;

    }

}bit[5005];

#define maxn 5005

const int inf=0x3f3f3f3f;

int wa[maxn],wb[maxn],wn[maxn],wv[maxn];

int rk[maxn],height[maxn],sa[maxn],r[maxn],Min[maxn][20],ok[maxn][maxn],n;

int cmp(int *r,int a,int b,int l)

{

    return (r[a]==r[b])&&(r[a+l]==r[b+l]);

}

void da(int *r,int *sa,int n,int m)

{

    int i,j,p,*x=wa,*y=wb,*t;

    for(i=0;i<m;i++) wn[i]=0;

    for(i=0;i<n;i++) wn[x[i]=r[i]]++;

    for(i=1;i<m;i++) wn[i]+=wn[i-1];

    for(i=n-1;i>=0;i--) sa[--wn[x[i]]]=i;

    for(j=1,p=1;p<n;j*=2,m=p)

    {

        for(p=0,i=n-j;i<n;i++) y[p++]=i;

        for(i=0;i<n;i++) if(sa[i]>=j) y[p++]=sa[i]-j;

        for(i=0;i<n;i++) wv[i]=x[y[i]];

        for(i=0;i<m;i++) wn[i]=0;

        for(i=0;i<n;i++) wn[wv[i]]++;

        for(i=1;i<m;i++) wn[i]+=wn[i-1];

        for(i=n-1;i>=0;i--) sa[--wn[wv[i]]]=y[i];

        for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1;i<n;i++)

            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;

    }

}

void calheight(int *r,int *sa,int n)

{

    int i,j,k=0;

    for(i=1;i<=n;i++) rk[sa[i]]=i;

    for(i=0;i<n;height[rk[i++]]=k )

    for(k?k--:0,j=sa[rk[i]-1];r[i+k]==r[j+k];k++);

}

void makermq()

{

    for(int i=1;i<=n;i++) Min[i][0]=height[i];

    for(int i=1;(1<<i)<=n;i++)

        for(int j=1;j+(1<<i)-1<=n;j++)

        {

            Min[j][i]=min(Min[j][i-1],Min[j+(1<<i-1)][i-1]);

        }

}

int ask(int a,int b)

{

    int l=rk[a],r=rk[b];

    if(l>r) swap(l,r);

    l++;

    if(l>r) return n-a;

    int tmp=int(log(r-l+1)/log(2));

    return min(Min[l][tmp],Min[r-(1<<tmp)+1][tmp]);

}

int check(int r,int l,int r1,int l1)

{

    r--,l--,r1--,l1--;

    if(r<0||l<0||r1<0||l1<0)return 0;

    if(ok[l1][r]==1)return 1;

    return 0;

}

long long updata(long long a,long long b)

{

    return (a+b)%mod;

}

int main()

{

    scanf("%d%s",&n,s+1);

    for(int i=0;i<n;i++)

        str[i]=s[i+1];

    for(int i=0;i<n;i++)

        r[i]=str[i];

    r[n]=0;

    da(r,sa,n+1,256);

    calheight(r,sa,n);

    makermq();

    for(int i = 0 ; i <= n ; ++ i) bit[i].init(n);

    for(int i = 0 ; i < n ; ++ i)

        for(int j = i + 1 ; j < n ; ++ j)

           if((j-i)%2==1){

                int tmp=ask(i,i+(j-i+1)/2);

                if(i+tmp>=i+(j-i+1)/2||str[i+tmp]>=str[i+(j-i+1)/2+tmp]) ok[i][j]=0;else ok[i][j]=1;

            }

    for(int i=1;i<=n;i++)

    {

        for(int j=1;j<=i;j++)

        {

            if(s[i-j+1] == '0')

                continue;

            dp[i][j] = 0 ;

            if(i-j == 0) dp[i][j] ++ ;

            dp[i][j] += bit[i-j].query(j - 1);

            if(i-j!=0&&(i-j-j+1)>0){

                if(ok[i-j-j][i-1])

                dp[i][j] += bit[i-j].query(j)-bit[i-j].query(j-1);

            }

            if(dp[i][j]>=mod)dp[i][j]%=mod;

            bit[i].updata(j,dp[i][j]);

        }

    }

    cout<<bit[n].query(n)<<endl;

    return 0;

}

巴特西

Codeforces Good Bye 2015 D. New Year and Ancient Prophecy 后缀数组树状数组 dp

D. New Year and Ancient Prophecy

题目连接：

Description

Input

Output

Sample Input

Sample Output

Hint

题意：

题解

代码

最新文章

热门文章

巴特西

Codeforces Good Bye 2015 D. New Year and Ancient Prophecy 后缀数组 树状数组 dp

D. New Year and Ancient Prophecy

题目连接：

Description

Input

Output

Sample Input

Sample Output

Hint

题意：

题解

代码

最新文章

热门文章

Codeforces Good Bye 2015 D. New Year and Ancient Prophecy 后缀数组树状数组 dp