2017 ACM-ICPC 亚洲区（西安赛区）网络赛 Barty's Computer(暴力+hash)

Barty have a computer, it can do these two things.

1. Add a new string to its memory, the length of this string is even.

2. For given $4$ strings $a,b,c,d$, find out how many strings that can be product by $a+s1+b+c+s2+d$, and $|a|+|s1|+|b|=|c|+|s2|+|d|$. $|s|$ means the length of string $s$, $s1$ and $s2$ can be any string, including "".

Please help your computer to do these things.

Input Format

Test cases begins with $T(T\le 5)$.

Then $T$ test cases follows.

Each test case begins with an integer $Q(Q\le 30000)$.

Then $Q$ lines,

1 s: add a new string $s$ to its memory.

2 a b c d: find how many strings satisfying the requirement above.

$\sum |s|+|a|+|b|+|c|+|d|\le 2000000$.

Output Format

For type $2$ query. Output the answer in one line.

样例输入

1101 abcqaq1 abcabcqaqqaq2 ab bc qa aq2 a c q q1 abcabcqaqqwq2 ab bc qa aq2 a c q q1 abcq2 a c q q2 a b c q

样例输出

121331

解：因为题目没有明确给出字符串的个数与长度 所以这里我们开一个长整型指针数组避免越界

一开始逐个字符匹配T了 所以加了hash预处理

坑点 并不是文本串的总长度大于4个字符串的总长度就可以 需要前两个 和 后两个字符串的长度总和小于等于 文本串的一半

#include <iostream>#include <stdio.h>#include <string.h>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <math.h>#include <bitset>#include <algorithm>#include <climits>using namespace std;const int N = 1e6+10;typedef long long LL;char str[N], a[N], b[N], c[N], d[N];const LL mod = 1e9+7;const LL seed = 1313;LL *hash1[N], pos[N];int xlen[N];int main(){    pos[0]=1;    for(int i=1;i<N;i++) pos[i]=pos[i-1]*seed%mod;    int t;    scanf("%d", &t);    while(t--)    {        int n;        scanf("%d",&n);        int h=0;        while(n--)        {            int xx;            scanf("%d", &xx);            if(xx==1)            {                scanf("%s",str+1);                int l=strlen(str+1);                xlen[h]=l;                hash1[h]=new LL[l+3];                hash1[h][0]=0;                for(int i=1;i<=l;i++)                    hash1[h][i]=(hash1[h][i-1]*seed%mod+(str[i]-'a'+1))%mod;                h++;            }            else            {                scanf("%s %s %s %s",a+1,b+1,c+1,d+1);                int l1=strlen(a+1),l2=strlen(b+1),l3=strlen(c+1),l4=strlen(d+1);                int sum=l1+l2+l3+l4, cnt=0;                LL ans1=0, ans2=0, ans3=0, ans4=0;                for(int j=1; j<=l1; j++) ans1=(ans1*seed%mod+(a[j]-'a'+1))%mod;                for(int j=1; j<=l2; j++) ans2=(ans2*seed%mod+(b[j]-'a'+1))%mod;                for(int j=1; j<=l3; j++) ans3=(ans3*seed%mod+(c[j]-'a'+1))%mod;                for(int j=1; j<=l4; j++) ans4=(ans4*seed%mod+(d[j]-'a'+1))%mod;                for(int i=0;i<h;i++)                {                    if((xlen[i]&1)||l1+l2>xlen[i]/2||l3+l4>xlen[i]/2) continue;                    int cx=xlen[i]/2;                    LL x=hash1[i][l1];                    if(x!=ans1) continue;                    x=(hash1[i][cx]-hash1[i][cx-l2]*pos[l2]%mod+mod)%mod;                    if(x!=ans2) continue;                    x=(hash1[i][cx+l3]-hash1[i][cx]*pos[l3]%mod+mod)%mod;                    if(x!=ans3) continue;                    x=(hash1[i][2*cx]-hash1[i][2*cx-l4]*pos[l4]%mod+mod)%mod;                    if(x!=ans4) continue;                    cnt++;                }                printf("%d\n",cnt);            }        }    }    return 0;}