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Another Meaning
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 678 Accepted Submission(s): 314
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.
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|
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
Hint
In the first case, “ hehehe” can have 3 meaings: “*he”, “he*”, “hehehe”. In the third case, “hehehehe” can have 5 meaings: “*hehe”, “he*he”, “hehe*”, “**”, “hehehehe”.
Author
FZU
Source
题意:有两个串A和B,其中B有两个意思,求A能组成多少种意思
思路,先用kmp把A串中相同的子串末尾标记起来,在来进行dp;
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int N=100005;
const int mod = 1e9+7;
int nexta[N],vis[N],dp[N];
char a[N],b[N];
int n,m;
void getnext()
{
memset(nexta,0,sizeof nexta);
int j=-1,i=0;
nexta[0]=-1;
while(i<n)
{
if(j==-1||b[j]==b[i]) i++,j++,nexta[i]=j;
else
j=nexta[j];
}
}
void kmp()
{
int i=0,j=0;
while(i<n&&j<m)
{
if(j==-1||a[i]==b[j])
i++,j++;
else
j=nexta[j];
if(j==m)
{
j=nexta[j];
vis[i]=1;
}
}
}
int main()
{
int t,ca=1;
scanf("%d",&t);
while(t--)
{
memset(vis,0,sizeof vis);
cin>>a>>b;
n=strlen(a);
m=strlen(b);
getnext();
kmp();
dp[0]=1;
for(int i=1;i<=n;i++)
{
dp[i]=dp[i-1];
if(vis[i])
dp[i]+=dp[i-m];
dp[i]%=mod;
}
printf("Case #%d: %d\n",ca++,dp[n]);
}
return 0;
}