3942 – Remember the WordTime limit: 3.000 seconds
Neal is very curious about combinatorial problems, and now here comes a problem about words. Knowing that Ray has a photographic memory and this may not trouble him, Neal gives it to Jiejie.
Since Jiejie can’t remember numbers clearly, he just uses sticks to help himself. Allowing for Jiejie’s only 20071027 sticks, he can only record the remainders of the numbers divided by total amount of sticks.
The problem is as follows: a word needs to be divided into small pieces in such a way that each piece is from some given set of words. Given a word and the set of words, Jiejie should calculate the number of ways the given word can be divided, using the words in the set.
Input
The input file contains multiple test cases. For each test case: the first line contains the given word whose length is no more than 300 000.
The second line contains an integer S , 1S4000 .
Each of the following S lines contains one word from the set. Each word will be at most 100 characters long. There will be no two identical words and all letters in the words will be lowercase.
There is a blank line between consecutive test cases.
You should proceed to the end of file.
Output
For each test case, output the number, as described above, from the task description modulo 20071027.
Sample Inputabcd 4 a b cd abSample OutputCase 1: 2
可以用Trie存字符串，，dp
#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<ctime>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i–)#define RepD(i,n) for(int i=n;i>=0;i–)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p]) #define Lson (x<<1)#define Rson ((x<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (20071027)#define MAXNode (100000*5+10)#define Sigma_size (26)#define MAXN (4000+10)#define MAXLen (300000+10)long long mul(long long a,long long b){return (a*b)%F;}long long add(long long a,long long b){return (a+b)%F;}long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}typedef long long ll;void upd(ll &a,ll b){a=(a%F+b%F)%F;}class Trie{public:int ch[MAXNode][Sigma_size];int v[MAXNode],siz;Trie(int _siz=0):siz(_siz){MEM(ch) MEM(v)}void mem(int _siz=0){siz=_siz; MEM(ch) MEM(v)}int idx(char c){return c-'a';}void insert(char *s,int val=0){int u=0,n=strlen(s);Rep(i,n){int c=idx(s[i]);if (!ch[u][c]){++siz;MEM(ch[siz]);ch[u][c]=siz;}u=ch[u][c];}v[u]=val;}void find(char *s,ll *x){int u=0,n=strlen(s);Rep(i,n){int c=idx(s[i]);if (!ch[u][c]){return;}u=ch[u][c];if (v[u]) upd(*x,*(x+i+1));}}}T;char s[MAXLen],t[MAXN];int n;ll d[MAXLen];int main(){//freopen("la3942.in","r",stdin);//freopen(".out","w",stdout);int tt=1;while(scanf("%s",s)==1){int len=strlen(s);scanf("%d",&n);T.mem(0);For(i,n) {scanf("%s",t);T.insert(t,1);}MEM(d) d[len]=1;RepD(i,len-1){T.find(s+i,d+i);}//Rep(i,len) cout<<d[i]<<' ';printf("Case %d: %d\n",tt++,d[0]);}return 0;}
从哪里跌倒就会从哪里爬起来,让我们一起努力吧
