题目链接:
Another Meaning
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
Problem Description
As is known to all, in many cases, a word has two meanings. Such as “hehe”, which not only means “hehe”, but also means “excuse me”.
Today, ?? is chating with MeiZi online, MeiZi sends a sentence A to ??. ?? is so smart that he knows the word B in the sentence has two meanings. He wants to know how many kinds of meanings MeiZi can express.
Today, ?? is chating with MeiZi online, MeiZi sends a sentence A to ??. ?? is so smart that he knows the word B in the sentence has two meanings. He wants to know how many kinds of meanings MeiZi can express.
Input
The first line of the input gives the number of test cases T; T test cases follow.
Each test case contains two strings A and B, A means the sentence MeiZi sends to ??, B means the word B which has two menaings. string only contains lowercase letters.
Limits
T <= 30
|A| <= 100000
|B| <= |A|
Each test case contains two strings A and B, A means the sentence MeiZi sends to ??, B means the word B which has two menaings. string only contains lowercase letters.
Limits
T <= 30
|A| <= 100000
|B| <= |A|
Output
For each test case, output one line containing “Case #x: y” (without quotes) , where x is the test case number (starting from 1) and y is the number of the different meaning of this sentence may be. Since this number may be quite large, you should output the answer modulo 1000000007.
Sample Input
4
hehehe
hehe
woquxizaolehehe
woquxizaole
hehehehe
hehe
owoadiuhzgneninougur
iehiehieh
Sample Output
Case #1: 3
Case #2: 2
Case #3: 5
Case #4: 1
题意:
给了一个单词它有两种意思,现在有一个字符串.问你这个字符串有多少种不同的意思;
思路:
设dp[i]表示[1,i]这段字符串的意思总数,考虑以i位结尾的的这一小段是否和单词匹配;
如果匹配,dp[i]=dp[i-1]+dp[i-strlen(B)];不匹配的话dp[i]=dp[i-1];
然后就是kmp找出所有的匹配位置的结点flag[i]标记下来;
AC代码:
/************************************************ ┆ ┏┓ ┏┓ ┆ ┆┏┛┻━━━┛┻┓ ┆ ┆┃ ┃ ┆ ┆┃ ━ ┃ ┆ ┆┃ ┳┛ ┗┳ ┃ ┆ ┆┃ ┃ ┆ ┆┃ ┻ ┃ ┆ ┆┗━┓ ┏━┛ ┆ ┆ ┃ ┃ ┆ ┆ ┃ ┗━━━┓ ┆ ┆ ┃ AC代马 ┣┓┆ ┆ ┃ ┏┛┆ ┆ ┗┓┓┏━┳┓┏┛ ┆ ┆ ┃┫┫ ┃┫┫ ┆ ┆ ┗┻┛ ┗┻┛ ┆ ************************************************ */ #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <bits/stdc++.h> #include <stack> using namespace std; #define For(i,j,n) for(int i=j;i<=n;i++) #define mst(ss,b) memset(ss,b,sizeof(ss)); typedef long long LL; template<class T> void read(T&num) { char CH; bool F=false; for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar()); for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar()); F && (num=-num); } int stk[70], tp; template<class T> inline void print(T p) { if(!p) { puts("0"); return; } while(p) stk[++ tp] = p%10, p/=10; while(tp) putchar(stk[tp--] + '0'); putchar(' '); } const LL mod=1e9+7; const double PI=acos(-1.0); const int inf=1e9; const int N=1e5+100; const int maxn=(1<<8); const double eps=1e-8; char A[N],B[N]; int nex[N],flag[N]; LL dp[N]; inline void makenext() { int k = -1,j = 0,len = strlen(B); nex[0] = -1; while(j < len) { if(k == -1||B[j] == B[k]) { j++; k++; nex[j] = k;//相等的话就往后继续; } else k = nex[k];//不等的话就相当于kmp一样,把模式串的这个子串用已经求出来的next跳转; } } inline void kmp() { mst(flag,0); makenext(); int posP = 0,posT = 0; int lenP = strlen(B),lenT = strlen(A); while(posP < lenP&&posT < lenT) { if(posP == -1||B[posP] == A[posT]) { posP++; posT++; } else posP = nex[posP]; if(posP == lenP) { flag[posT-1]=1; posP=nex[posP]; } } } int main() { int t,Case=0; read(t); while(t--) { scanf("%s",A); scanf("%s",B); kmp(); int len=strlen(A),l=strlen(B); dp[0]=1; For(i,1,len) { if(!flag[i-1])dp[i]=dp[i-1]; else dp[i]=(dp[i-1]+dp[i-l])%mod; } printf("Case #%d: %lld ",++Case,dp[len]); } return 0; }