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=0;register char c=getchar();for(;!isdigit(c);c=getchar()); for(;isdigit(c);now=now*10+c-'0',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 ", x) #define Prl(x) printf("%lld ",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-9; const int maxn = 1e6 + 10; const int INF = 0x3f3f3f3f; const LL mod = 998244353 ; 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[0] = m; int j = 0; while(j + 1 < m && x[j] == x[j + 1]) j++; next[1] = j; int k = 1; for(int i = 2; i < m ; i ++){ int p = next[k] + k - 1; int L = next[i - k]; if(i + L < p + 1) next[i] = L; else{ j = max(0,p - i + 1); 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 = 0; while(j < n && j < m && x[j] == y[j]) j ++; extend[0] = j; int k = 0; for(int i = 1; i < n ; i ++){ int p = extend[k] + k - 1; int L = next[i - k]; if(i + L < p + 1) extend[i] = L; else{ j = max(0,p - i + 1); 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 << 3]; void Build(int t,int l,int r){ if(l > r) exit(0); tree[t].l = l; tree[t].r = r; tree[t].lazy = 0; if(l == r) return; int m = (l + r) >> 1; Build(t << 1,l,m); Build(t << 1 | 1,m + 1,r); } void Pushdown(int t){ if(tree[t].lazy){ tree[t << 1].lazy = (tree[t << 1].lazy + tree[t].lazy) % mod; tree[t << 1 | 1].lazy = (tree[t << 1 | 1].lazy + tree[t].lazy) % mod; tree[t].lazy = 0; } } 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) >> 1; if(r <= m) update(t << 1,l,r,v); else if(l > m) update(t << 1 | 1,l,r,v); else{ update(t << 1,l,m,v); update(t << 1 | 1,m + 1,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) >> 1; if(p <= m) return query(t << 1,p); else return query(t << 1 | 1,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 = 1; i <= N ; i ++){ l[i] = i + l1 - 1,r[i] = i + l2 - 1; if(a[i - 1] == '0'){ if(down[0] == '0'){ l[i] = r[i] = i; }else{ l[i] = 1,r[i] = 0; } continue; } int len = edown[i - 1]; if((len < l1) && down[len] > a[i + len - 1]) l[i]++; len = eup[i - 1]; if((len < l2) && up[len] < a[i + len - 1]) r[i]--; } Build(1,0,N * 2); update(1,0,0,1); for(int i = 1; i <= N ; i ++){ if(l[i] > r[i]) continue; LL x = query(1,i - 1); update(1,l[i],r[i],x); } Prl(query(1,N)); return 0; }