【codeforces #282(div 1)】AB题解

#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <cmath> #define M 100000+5 using namespace std; char s[M]; int ans[M]; int main() { scanf("%s",s); int l=strlen(s); int la,now=0,tot=0; for (int i=0;i<l;i++) { if (s[i]=='#') { ans[++tot]=1; now--; la=i; } if (s[i]=='(') now++; if (s[i]==')') now--; if (now<0) { puts("-1"); return 0; } } int x=0; for (int i=l-1;i>la;i--) { if (s[i]=='(') x--; else x++; if (x<0) { puts("-1"); return 0; } } ans[tot]+=now; for (int i=1;i<=tot;i++) printf("%d\n",ans[i]); return 0; }

Hamed has recently found a string t and suddenly became quite fond of it. He spent several days trying to find all occurrences of t in other strings he had. Finally he became tired and started thinking about the following problem. Given a string s how many ways are there to extract k?≥?1 non-overlapping substrings from it such that each of them contains string t as a substring? More formally, you need to calculate the number of ways to choose two sequences a1,?a2,?...,?ak and b1,?b2,?...,?bk satisfying the following requirements:

k?≥?1
$技术分享$
$技术分享$
$技术分享$
$技术分享$ t is a substring of string saisai?+?1... sbi (string s is considered as 1-indexed).

As the number of ways can be rather large print it modulo 109?+?7.

Input

Input consists of two lines containing strings s and t (1?≤?|s|,?|t|?≤?105). Each string consists of lowercase Latin letters.

Output

Print the answer in a single line.

Sample test(s)

input
ababa
aba

output
5

input
welcometoroundtwohundredandeightytwo
d

output
274201

input
ddd
d

output
12

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cstdlib> #define mod 1000000007 #define M 100000+5 using namespace std; int ss[M],ne[M],n,m,ok[M],sum[M],f[M]; char s[M],t[M]; void Getfail() { ne[0]=0; ne[1]=0; for (int i=1;i<m;i++) { int j=ne[i]; while (j&&t[i]!=t[j]) j=ne[j]; ne[i+1]=t[i]==t[j]?j+1:0; } } void Find() { Getfail(); int j=0; int now=0; for (int i=0;i<n;i++) ok[i]=n; for (int i=0;i<n;i++) { while (j&&t[j]!=s[i]) j=ne[j]; if (t[j]==s[i]) j++; if (j==m) { for (int k=now;k<=i-m+1;k++) ok[k]=min(i,ok[k]); now=i-m+2; j=ne[j]; } } } int main() { scanf("%s",s); scanf("%s",t); n=strlen(s),m=strlen(t); Find(); for (int i=n-1;i>=0;i--) { f[i]=n-1-(ok[i]-1); f[i]=(f[i]+ss[ok[i]+1])%mod; sum[i]=(sum[i+1]+f[i])%mod; ss[i]=(ss[i+1]+sum[i])%mod; } cout<<sum[0]%mod<<endl; return 0; }