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 << "\n";
#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 '\n'

#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;  //
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //
const int mod = ;
const double esp = 1e-;
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=;int f=;char ch=getchar();
    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
    return x=f?-x:x;
}
/*-----------------------showtime----------------------*/
            const int maxn = 1e6+;
            const int base = ;
            const int base2 = ;
            const int MOD = 1e9+;
            //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[]=val1[]=;
                    qp1[]=p1[]=;
                    len=strlen(s+);
                    for(int i=;i<=len;i++)
                    {
                        qp1[i]=(qp1[i-]*base)%MOD;
                        h1[i]=(h1[i-]*base%MOD+s[i])%MOD;
                    }

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

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

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

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

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

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

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

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

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

            }

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

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

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