hdu6153（kmp） A Secret

A Secret

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 256000/256000 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

Today is the birthday of SF,so VS gives two strings S1,S2 to SF as a present,which have a big secret.SF is interested in this secret and ask VS how to get it.There are the things that VS tell:
  Suffix(S2,i) = S2[i...len].Ni is the times that Suffix(S2,i) occurs in S1 and Li is the length of Suffix(S2,i).Then the secret is the sum of the product of Ni and Li.
  Now SF wants you to help him find the secret.The answer may be very large, so the answer should mod 1000000007.

 

Input

Input contains multiple cases.
  The first line contains an integer T,the number of cases.Then following T cases.
  Each test case contains two lines.The first line contains a string S1.The second line contains a string S2.
  1<=T<=10.1<=|S1|,|S2|<=1e6.S1 and S2 only consist of lowercase ,uppercase letter.

 

Output

For each test case,output a single line containing a integer,the answer of test case.
  The answer may be very large, so the answer should mod 1e9+7.

 

Sample Input

2aaaaaaaabababababa

 

Sample Output

1319
Hint
case 2: Suffix(S2,1) = "aba",Suffix(S2,2) = "ba",Suffix(S2,3) = "a".N1 = 3,N2 = 3,N3 = 4.L1 = 3,L2 = 2,L3 = 1.ans = （3*3+3*2+4*1）%1000000007.【大佬】的题解，借大佬的题解补一下题目 
#include<cstdio>#include<iostream>#include<cstring>using namespace std;#define ll long longconst int mn=1e6+5;int n,m,nextval[mn];char s[mn],t[mn];ll id[mn];const int mod=1e9+7;void getNextval() {    int i=0,j=-1;    nextval[0]=-1;    while(i<m) {        if(j==-1||t[i]==t[j])            nextval[++i]=++j;        else j=nextval[j];    }}void KMP() {    int i=0,j=0;    while(i<n) {        if(j==-1||s[i]==t[j]) ++i,++j;        else j=nextval[j];        id[j]++;        if(j==m)        j=nextval[j];    }}int main() {    int T;    scanf("%d",&T);    while(T--) {        memset(id,0,sizeof(id));        scanf("%s%s",s,t);        n=strlen(s),m=strlen(t);        for(int i=0;i<=(n-1)/2;++i)        swap(s[i],s[n-1-i]);        for(int i=0;i<=(m-1)/2;++i)        swap(t[i],t[m-1-i]);        getNextval();        KMP();        ll ans=0;        for(int i=m;i>0;--i)        {        id[nextval[i]]+=id[i];        ans=(ans+id[i]*i)%mod;}        printf("%lld\n",ans);    }    return 0;}

【KMP的原样】：（我自己搜出来的，请指教啊）
#include <stack>#include <vector>#include <set>#include <map>#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<string>using namespace std;#define MAX_SIZE 100010typedef char ElemType;void CptPfFunc(ElemType Pattern[], int PrefixFunc[]);void KMPstrMatching(ElemType Target[], ElemType Pattern[]);char p[MAX_SIZE];char t[MAX_SIZE];int cnt;            ///记录匹配的的个数int main(){    int re;    scanf("%d",&re);    while(re--){        printf("please input string t:");        cin>>t;//        t.length=strlen(t.string);        printf("please input string p:");        cin>>p;//        p.length=strlen(p.string);//        getnext(p,next);//        printf("\n%d\n",kmp(t,p,next));        cnt=0;        KMPstrMatching(t,p);    }    return 0;}/// Compute Prefix functionvoid CptPfFunc( ElemType Pattern[], int PrefixFunc[] ){    register int iLen = 0;    /// Length of Pattern[]    while('\0'!=Pattern[iLen])            iLen++;    int LOLP=0;     /// Lenth of longest prefix    PrefixFunc[1]=0;    for(int NOCM=2;NOCM<iLen+1;NOCM++)     /// NOCM represent the number of characters matched    {            while(LOLP>0&&(Pattern[LOLP]!=Pattern[NOCM-1]))                    LOLP=PrefixFunc[LOLP];            if(Pattern[LOLP]==Pattern[NOCM-1])                    LOLP++;            PrefixFunc[NOCM]=LOLP;    }}void KMPstrMatching(ElemType Target[],ElemType Pattern[]){    int PrefixFunc[MAX_SIZE];    register int TarLen=0;    register int PatLen=0;    /// Compute the length of array Target and Pattern    while('\0'!=Target[TarLen])            TarLen++;    while('\0'!=Pattern[PatLen])            PatLen++;    /// Compute the prefix function of Pattern    CptPfFunc(Pattern,PrefixFunc);    int NOCM=0;     /// Number of characters matched    for(int i=0;i<TarLen;i++ )    {        while(NOCM>0&&Pattern[NOCM]!=Target[i])                NOCM=PrefixFunc[NOCM];        if(Pattern[NOCM]==Target[i])                NOCM++;        if(NOCM==PatLen )        {                cout<<"KMP String Matching,pattern occurs with shift "<<i-PatLen+1<<endl;                cnt++;                NOCM=PrefixFunc[NOCM];        }    }    cout<<"cnt=: "<<cnt<<endl;}