CF_EDU51 E. Vasya and Big Integers

传送门：https://codeforces.com/contest/1051/problem/E

题意：

　　把一个数分成许多小段，每一段的值在L和R间。问有多少种分法。

思路 ：

　　首先，需要快速处理出每个位子 ，最近和最远可以组合的区间。如何处理这个区间，由于时限比较紧张，可以用哈希的方法，因为是长度固定，所以可以用二分确定两端长度前缀，再比较大小的方法。当然哈希要用比较好的双哈希。

　　然后，我们可以从后往前dp，每个位子的答案 = 这个位子对应区间的值的和。

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "
";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '
'

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c);
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 998244353;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;


template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}
/*-----------------------showtime----------------------*/
            const int maxn = 1e6+9;
            const int base = 37;
            const int base2 = 71;
            const int MOD = 1e9+7;
            //const ll MOD2 = 1e16+17;
            int dple[maxn],dpri[maxn];
            int lenle,lenri;
            ll dp[maxn],sum[maxn];
            struct hash
            {
                char s[maxn];
                ll h1[maxn],qp1[maxn];
                ull h2[maxn],qp2[maxn];
                ll p1[maxn],p2[maxn],len,val1[maxn],val2[maxn];

                void init()
                {
                    h1[0]=val1[0]=0;
                    qp1[0]=p1[0]=1;
                    len=strlen(s+1);
                    for(int i=1;i<=len;i++)
                    {
                        qp1[i]=(qp1[i-1]*base)%MOD;
                        h1[i]=(h1[i-1]*base%MOD+s[i])%MOD;
                    }

                    h2[0]=val2[0]=0;
                    qp2[0]=p2[0]=1;
                    for(int i=1;i<=len;i++)
                    {
                        qp2[i]=(qp2[i-1]*base2);
                        h2[i]=(h2[i-1]*base2+s[i]);
                    }
                }
                ll get_hash1(int l,int r)
                {
                    return ((h1[r]-h1[l-1]*qp1[r-l+1])%MOD +MOD )% MOD;
                }
                ull get_hash2(int l,int r)
                {
                    return ((h2[r]-h2[l-1]*qp2[r-l+1]));
                }
            }A,L,R;

            bool checkle(int l1, int r1) {
                    int le = l1, ri = r1;
                    while(le <= ri){
                        int mid = (le + ri) >> 1;
                        if(A.get_hash1(le,mid) == L.get_hash1(le-l1+1,mid-l1+1)  && A.get_hash2(le,mid) == L.get_hash2(le-l1+1,mid-l1+1)){
                            le = mid+1;
                        }
                        else ri = mid - 1;
                    }
                    if(le == r1 + 1) return true;
                    return A.s[le] >= L.s[le - l1 + 1];
            }

            bool checkri(int l1,int r1) {
                    int le = l1, ri = r1;

                    while(le <= ri){
                        int mid = (le + ri) >> 1;

                        if(A.get_hash1(le,mid) == R.get_hash1(le-l1+1,mid-l1+1) &&A.get_hash2(le,mid) == R.get_hash2(le-l1+1,mid-l1+1) ){
                            le = mid+1;
                        }
                        else ri = mid-1;
                    }
                    if(le == r1 + 1) return true;
                    return A.s[le] <= R.s[le - l1 + 1];
            }


int main(){
            scanf("%s%s%s", A.s + 1, L.s + 1,R.s + 1);
            A.init();   L.init();   R.init();

            ll len = A.len;
            int n = len;
            lenle = L.len; lenri = R.len;
            for(int i=1; i<=n; i++){
                int le = i, ri = n;
                dple[i] = -1; dpri[i] = -1;
                if(A.s[i] == '0') {
                        if(L.s[1] == '0') dple[i] = i,dpri[i] = i;
                        else dple[i] = -1,dpri[i] = -1;
                        continue;
                }

                if(i + lenle - 1<= n &&  checkle(i, i + lenle - 1)) dple[i] = i + lenle - 1;
                else if(i + lenle<=n)dple[i] = i + lenle;


                if(i + lenri - 1<= n&& checkri(i, i + lenri -1)) dpri[i] = i + lenri - 1;
                else dpri[i] = min(n,i+lenri-2);

            }

            sum[n+1] = 1;
            for(int  i=n; i>=1; i--){
                int pl = dple[i] + 1,pr = dpri[i] + 1;

                dp[i] = ((sum[pl] - sum[pr + 1] ) % mod + mod )%mod;
                sum[i] = (sum[i+1] + dp[i]) % mod;
            }

            cout<<dp[1]<<endl;
            return 0;
}