2017西安邀请赛: I. 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

题目链接：https://nanti.jisuanke.com/t/17122

Q只有30000，总长也不超过2000000

上去就是一阵暴力，如果T了，说明只是需要一点小小的Hash

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define mod 73939133#define LL long longLL h1, h2, h3, h4, *Hash[30005], Pow[2000005] = {1};int l1, l2, l3, l4, cnt, len[30005];char *str[30005], temp[2000005], a[2000005], b[2000005], c[2000005], d[2000005];LL Jud(int id, int x, int y){return (Hash[id][y]-Hash[id][x-1]%mod*Pow[y-x+1]%mod+mod)%mod;}int main(void){int T, opt, q, i, sum;for(i=1;i<=2000005;i++)Pow[i] = Pow[i-1]*128%mod;scanf("%d", &T);while(T--){cnt = 0;scanf("%d", &q);while(q--){scanf("%d", &opt);if(opt==1){scanf("%s", temp+1);len[++cnt] = strlen(temp+1);str[cnt] = new char[len[cnt]+3];for(i=1;i<=len[cnt];i++)str[cnt][i] = temp[i];str[len[cnt]+1] = 0;Hash[cnt] = new LL[len[cnt]+3];Hash[cnt][0] = 0;for(i=1;i<=len[cnt];i++)Hash[cnt][i] = (Hash[cnt][i-1]*128+temp[i])%mod;}else{scanf("%s%s%s%s", a+1, b+1, c+1, d+1);l1 = strlen(a+1);l2 = strlen(b+1);l3 = strlen(c+1);l4 = strlen(d+1);h1 = h2 = h3 = h4 = 0;for(i=1;i<=l1;i++)h1 = (h1*128+a[i])%mod;for(i=1;i<=l2;i++)h2 = (h2*128+b[i])%mod;for(i=1;i<=l3;i++)h3 = (h3*128+c[i])%mod;for(i=1;i<=l4;i++)h4 = (h4*128+d[i])%mod;//printf("%lld %lld %lld %lld\n", h1, h2, h3, h4);sum = 0;for(i=1;i<=cnt;i++){if(l1+l2>len[i]/2 || l3+l4>len[i]/2)continue;//printf("%lld %lld %lld\n", Hash[1][2], Hash[1][3], Hash[1][3]-Hash[1][2]*Pow[1]);//printf("%lld %d %d\n", Jud(i, len[i]/2-l2+1, len[i]/2), len[i]/2-l2+1, len[i]/2);if(Jud(i, 1, l1)==h1 && Jud(i, len[i]/2-l2+1, len[i]/2)==h2 && Jud(i, len[i]/2+1, len[i]/2+l3)==h3 && Jud(i, len[i]-l4+1, len[i])==h4)sum++;}printf("%d\n", sum);}}for(i=1;i<=cnt;i++){delete []str[i];delete []Hash[i];}}return 0;}/*1101 abcqaq2 a c q q*/