Codeforces 451 D. Count Good Substrings
2014-07-27 14:39
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D. Count Good Substrings
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
We call a string good, if after merging all the consecutive equal characters, the resulting string is palindrome. For example, "aabba"
is good, because after the merging step it will become "aba".
Given a string, you have to find two values:
the number of good substrings of even length;
the number of good substrings of odd length.
Input
The first line of the input contains a single string of length n (1 ≤ n ≤ 105).
Each character of the string will be either 'a' or 'b'.
Output
Print two space-separated integers: the number of good substrings of even length and the number of good substrings of odd length.
Sample test(s)
input
output
input
output
input
output
input
output
Note
In example 1, there are three good substrings ("b", "b", and
"bb"). One of them has even length and two of them have odd length.
In example 2, there are six good substrings (i.e. "b", "a",
"a", "b", "aa",
"baab"). Two of them have even length and four of them have odd length.
In example 3, there are seven good substrings (i.e. "b", "a",
"b", "b", "bb",
"bab", "babb"). Two of them have even length and five of them
have odd length.
Definitions
A substring s[l, r] (1 ≤ l ≤ r ≤ n) of
string s = s1s2... sn is
string slsl + 1... sr.
A string s = s1s2... sn is
a palindrome if it is equal to string snsn - 1... s1.
import java.util.*;
public class Main
{
public static void main(String[] args)
{
Scanner in=new Scanner(System.in);
String str=in.nextLine();
long a0=0,a1=0,b0=0,b1=0,d=0;
for(int i=0;i<str.length();i++)
{
if(str.charAt(i)=='a')
{
if(i%2==0)
{
a0++;
d+=a0;
}
else
{
a1++;
d+=a1;
}
}
else if(str.charAt(i)=='b')
{
if(i%2==0)
{
b0++;
d+=b0;
}
else
{
b1++;
d+=b1;
}
}
}
System.out.println((a0*a1+b0*b1)+" "+d);
}
}
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
We call a string good, if after merging all the consecutive equal characters, the resulting string is palindrome. For example, "aabba"
is good, because after the merging step it will become "aba".
Given a string, you have to find two values:
the number of good substrings of even length;
the number of good substrings of odd length.
Input
The first line of the input contains a single string of length n (1 ≤ n ≤ 105).
Each character of the string will be either 'a' or 'b'.
Output
Print two space-separated integers: the number of good substrings of even length and the number of good substrings of odd length.
Sample test(s)
input
bb
output
1 2
input
baab
output
2 4
input
babb
output
2 5
input
babaa
output
2 7
Note
In example 1, there are three good substrings ("b", "b", and
"bb"). One of them has even length and two of them have odd length.
In example 2, there are six good substrings (i.e. "b", "a",
"a", "b", "aa",
"baab"). Two of them have even length and four of them have odd length.
In example 3, there are seven good substrings (i.e. "b", "a",
"b", "b", "bb",
"bab", "babb"). Two of them have even length and five of them
have odd length.
Definitions
A substring s[l, r] (1 ≤ l ≤ r ≤ n) of
string s = s1s2... sn is
string slsl + 1... sr.
A string s = s1s2... sn is
a palindrome if it is equal to string snsn - 1... s1.
import java.util.*;
public class Main
{
public static void main(String[] args)
{
Scanner in=new Scanner(System.in);
String str=in.nextLine();
long a0=0,a1=0,b0=0,b1=0,d=0;
for(int i=0;i<str.length();i++)
{
if(str.charAt(i)=='a')
{
if(i%2==0)
{
a0++;
d+=a0;
}
else
{
a1++;
d+=a1;
}
}
else if(str.charAt(i)=='b')
{
if(i%2==0)
{
b0++;
d+=b0;
}
else
{
b1++;
d+=b1;
}
}
}
System.out.println((a0*a1+b0*b1)+" "+d);
}
}
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