【codeforces #282(div 1)】AB题解

A. Treasure

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Malek has recently found a treasure map. While he was looking for a treasure he found a locked door. There was a string s written on the door consisting
of characters '(', ')' and '#'.
Below there was a manual on how to open the door. After spending a long time Malek managed to decode the manual and found out that the goal is to replace each '#'
with one or more ')' characters so that the final string becomes beautiful.

Below there was also written that a string is called beautiful if for each i (1 ≤ i ≤ |s|)
there are no more ')' characters than '('
characters among the first i characters of s and
also the total number of '(' characters is equal to the total number of ')'
characters.

Help Malek open the door by telling him for each '#' character how many ')'
characters he must replace it with.

Input

The first line of the input contains a string s (1 ≤ |s| ≤ 105).
Each character of this string is one of the characters '(', ')'
or '#'. It is guaranteed that s contains
at least one '#' character.

Output

If there is no way of replacing '#' characters which leads to a beautiful string print  - 1.
Otherwise for each character '#' print a separate line containing a positive integer, the number of ')'
characters this character must be replaced with.

If there are several possible answers, you may output any of them.

Sample test(s)

input

(((#)((#)

output

1
2

input

()((#((#(#()

output

2
2
1

input

#

output

-1

input

(#)

output

-1

贪心。

(表示1，)表示-1。

满足条件则前缀和时刻都要>=0。

那么遇到#我们只让他表示一个)，前缀和只-1，遇到最后一个#再把前面的债还清。

一开始WA了，因为我遇到最后一个#就不管前缀和了。。

因此(#(这样的数据就过不了。

#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;
}

B. Obsessive String

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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








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

kmp+dp。

首先用kmp快速求出每一位的ok[i]，也就是从i到ok[i]包含t，且ok[i]最小。

然后进行dp:

f[i]表示a[1]=i的方案数，这显然要倒着做。

f[i]=sigma(sigma(f[ok[i]+1...n-1)+sigma(f[ok[i]+2...n-1]...+f[n-1]))

维护后缀和sum[i]表示i到n-1的f值得后缀和。

再维护后缀和的后缀和ss[i]表示i到n-1的sum[i]的后缀和。

于是f[i]=ss[ok[i]+1]，转移变成O(1)了！

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#define mod 1000000007
#define M 100000+5using 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;
}