http://codeforces.com/problemset/problem/1051/E
题意：给你一个很大的数字，然后你可以把这个数字拆分成为任意多个部分，要求每一个部分的数字大小要在一个区间内，问有多少种拆分方式。
很容易看出这是一个dp，用dp[i]表示到i之前位置总共的数量，再用l[i]和r[i]表示i位置到l和r区间内的字符串全都满足上下限的条件，将dp[i - 1]加到l到r上更新即可，是一个很显然的n²dp，当然n²是不可能的，随随便便搞个数据结构进行一下区间修改就行了，这里用的是线段树。
问题就给到了预处理l[i]和r[i]这个问题上，暴力预处理又是一个n²的操作，考虑到大数比较事实上是去掉最大公共前缀之后比较下一位即可，可以容易的想到用EXKMP去处理下就好了。
（线段树一开始就开了0 ~ N，RE了一个小时）
#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
inline int read(){int now=;register char c=getchar();for(;!isdigit(c);c=getchar());
for(;isdigit(c);now=now*+c-'',c=getchar());return now;}
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x);
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
const double eps = 1e-;
const int maxn = 1e6 + ;
const int INF = 0x3f3f3f3f;
const LL mod = ;
char a[maxn],down[maxn],up[maxn];
int l[maxn],r[maxn];
int nxt[maxn],edown[maxn],eup[maxn];
void pre_EKMP(char x[],int m,int next[]){
next[] = m;
int j = ;
while(j + < m && x[j] == x[j + ]) j++;
next[] = j;
int k = ;
for(int i = ; i < m ; i ++){
int p = next[k] + k - ;
int L = next[i - k];
if(i + L < p + ) next[i] = L;
else{
j = max(,p - i + );
while(i + j < m && x[i + j] == x[j]) j ++;
next[i] = j;
k = i;
}
}
}
void EKMP(char x[],int m,char y[],int n,int next[],int extend[]){
pre_EKMP(x,m,next);
int j = ;
while(j < n && j < m && x[j] == y[j]) j ++;
extend[] = j;
int k = ;
for(int i = ; i < n ; i ++){
int p = extend[k] + k - ;
int L = next[i - k];
if(i + L < p + ) extend[i] = L;
else{
j = max(,p - i + );
while(i + j < n && j < m && y[i + j] == x[j]) j ++;
extend[i] = j;
k = i;
}
}
}
struct Tree{
int l,r;
LL lazy;
}tree[maxn << ];
void Build(int t,int l,int r){
if(l > r) exit();
tree[t].l = l; tree[t].r = r;
tree[t].lazy = ;
if(l == r) return;
int m = (l + r) >> ;
Build(t << ,l,m); Build(t << | ,m + ,r);
}
void Pushdown(int t){
if(tree[t].lazy){
tree[t << ].lazy = (tree[t << ].lazy + tree[t].lazy) % mod;
tree[t << | ].lazy = (tree[t << | ].lazy + tree[t].lazy) % mod;
tree[t].lazy = ;
}
}
void update(int t,int l,int r,LL v){
if(l <= tree[t].l && tree[t].r <= r){
tree[t].lazy = (v + tree[t].lazy) % mod;
return;
}
Pushdown(t);
int m = (tree[t].l + tree[t].r) >> ;
if(r <= m) update(t << ,l,r,v);
else if(l > m) update(t << | ,l,r,v);
else{
update(t << ,l,m,v);
update(t << | ,m + ,r,v);
}
}
LL query(int t,int p){
if(tree[t].l >= tree[t].r) return tree[t].lazy % mod;
Pushdown(t);
int m = (tree[t].l + tree[t].r) >> ;
if(p <= m) return query(t << ,p);
else return query(t << | ,p);
}
int main()
{
scanf("%s%s%s",a,down,up);
int N = strlen(a);
int l1 = strlen(down),l2 = strlen(up);
EKMP(down,l1,a,N,nxt,edown);
EKMP(up,l2,a,N,nxt,eup);
for(int i = ; i <= N ; i ++){
l[i] = i + l1 - ,r[i] = i + l2 - ;
if(a[i - ] == ''){
if(down[] == ''){
l[i] = r[i] = i;
}else{
l[i] = ,r[i] = ;
}
continue;
}
int len = edown[i - ];
if((len < l1) && down[len] > a[i + len - ]) l[i]++;
len = eup[i - ];
if((len < l2) && up[len] < a[i + len - ]) r[i]--;
}
Build(,,N * ); update(,,,);
for(int i = ; i <= N ; i ++){
if(l[i] > r[i]) continue;
LL x = query(,i - );
update(,l[i],r[i],x);
}
Prl(query(,N));
return ;
}