HDU 6153 A secret(kmp)

似乎正解是扩展kmp的模版题.
但是kmp也可以做.
kmp有两种.一种是在每一次成功匹配后都算一次.
这种是正解呢.
因为kmp,其实就是遍历了所有会有重叠的情况,但是一些毫无可能的就被忽略了,但是我们只记录成功匹配一个之后的情况的话,会有在while中的匹配被忘记.所以没一个成功匹配到之后,都有他的next被忽略了.
一种是在结束匹配是算一次,这种要等差数列计算.
反正这题很迷幻.
靠我总结不下去了,还是不懂啊.
就当存个kmp,我好菜啊

/* Farewell. */#include <iostream>#include <vector>#include <cstdio>#include <string.h>#include <cstring>#include <algorithm>#include <queue>#include <map>#include <string>#include <cmath>#include <bitset>#include <iomanip>#include <set>using namespace std;#define gcd __gcd#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define MP make_pair#define MT make_tuple#define PB push_backtypedef long long  LL;typedef unsigned long long ULL;typedef pair<int,int > pii;typedef pair<LL,LL> pll;typedef pair<double,double > pdd;typedef pair<double,int > pdi;const int INF = 0x7fffffff;const LL INFF = 0x7f7f7f7fffffffff;#define debug(x) std::cerr << #x << " = " << (x) << std::endlconst int MAXM = 5e3+17;const int MOD = 1e9+7;const int MAXN = 1e6+17;char s[MAXN],p[MAXN];string tmp;LL Next[MAXN];LL arv[MAXN];void getNext(string p){    memset(Next, 0, sizeof(Next));    int m = p.length();    int i = 0,j = -1;    Next[0] = -1;    while (i < m)    {        if (j == -1 || p[i] == p[j])        {            i++;j++;            Next[i] = j;        }else j = Next[j];    }}void kmp(string s,string p){    memset(arv, 0, sizeof(arv));    int i = 0, j = 0,n = s.length(),m = p.length();    getNext(p);    for(int i =0 ;i<n;++i)    {        while(j > 0 && s[i] != p[j])        {            arv[j]++;            j = Next[j];        }        if(s[i] == p[j])            j++;        if(j == m)        {            arv[j]++;            j = Next[j];        }    }}int main(int argc, char const *argv[]){            #ifdef GoodbyeMonkeyKing    freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);    #endif    int t;    cin>>t;    while(t--)    {        scanf("%s%s",s,p);        string a(s,&s[strlen(s)]);        string b(p,&p[strlen(p)]);        reverse(a.begin(), a.end());        reverse(b.begin(), b.end());        kmp(a,b);        LL ans = 0;        for (LL i = 0; i <= b.length(); ++i)            ans = (ans+(1LL*arv[i]*(1LL*(i+1)*i)/2)%MOD)%MOD;        cout<<ans%MOD<<endl;    }    return 0;    }