CF451D Count Good Substrings （DP）

Codeforces Round #258 (Div. 2)

Count Good Substrings

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:

1. the number of good substrings of even length;
2. 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... s_n is string s_ls_l + 1... s_r.

A string s = s₁s₂... s_n is a palindrome if it is equal to string s_ns_n - 1... s₁.

题意：若一个字符串删除相同元素后剩下的是回文串，则称之为好串。给出一个由a和b组成的字符串，求奇数元素的好子串数量和偶数元素的好子串数量。

题解：DP统计。

注意字符串只有a和b，删除重复元素以后肯定是ababababab这样，两个a和之间的元素组成的串肯定是回文串，根本不用回文的算法。

由s[i]结尾的回文串数量等于0~i之间和s[i]相同字符的数量。要分奇数偶数长度，这和下标的奇偶有关。我们观察一下当前下标、之前的和当前字符相同的字符的下标、ans统计下标的关系（ans[0]记录偶数，ans[1]记录奇数）

now	之前	ans
奇	奇	奇
奇	偶	偶
偶	奇	偶
偶	偶	奇

可以发现当前下标为奇数，当前元素为a，ans偶+=之前的偶数位置的a的数量，ans奇+=之前奇数位置a的数量。

当前下标为偶数，当前元素为a，ans偶+=之前的奇数位置的a的数量，ans奇+=之前偶数位置a的数量。

这样就总结出了比较简单的统计方法。（其实不总结，直接用if也可以，比赛时最好用直接点的方法，我这是事后做的才搞这种）

这样我们从头扫到尾，慢慢统计字符的数量和奇数偶数回文串的数量就行了。

统计具体做法：

         for(i=; i<len; i++) {
             bool t=i&;///t=i%2
             ans[]+=f[!t][ID[s[i]]];
             ans[]+=f[ t][ID[s[i]]];
             ans[]++;
             f[t][ID[s[i]]]++;
         }

其中ID['a']=0，ID['b']=1，ans[0]统计偶数，ans[1]统计奇数，f[t][j]中t表示奇偶，j表示a或者b，f[][]统计之前的奇/偶位置的a/b的数量。

全代码：

 //#pragma comment(linker, "/STACK:102400000,102400000")
 #include<cstdio>
 #include<cmath>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<map>
 #include<set>
 #include<stack>
 #include<queue>
 using namespace std;
 #define ll long long
 #define usint unsigned int
 #define mz(array) memset(array, 0, sizeof(array))
 #define minf(array) memset(array, 0x3f, sizeof(array))
 #define REP(i,n) for(i=0;i<(n);i++)
 #define FOR(i,x,n) for(i=(x);i<=(n);i++)
 #define RD(x) scanf("%d",&x)
 #define RD2(x,y) scanf("%d%d",&x,&y)
 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
 #define WN(x) printf("%d\n",x);
 #define RE  freopen("D.in","r",stdin)
 #define WE  freopen("1biao.out","w",stdout)
 
 char s[];
 ll f[][];
 ll ans[];
 int len;
 int ID[];
 
 int main() {
     ID['a']=;
     ID['b']=;
     int i,j,k;
     while(gets(s)!=NULL) {
         len=strlen(s);
         mz(f);
         mz(ans);
         for(i=; i<len; i++) {
             bool t=i&;///t=i%2
             ans[]+=f[!t][ID[s[i]]];
             ans[]+=f[ t][ID[s[i]]];
             ans[]++;
             f[t][ID[s[i]]]++;
         }
         printf("%I64d %I64d\n",ans[],ans[]);
     }
     return ;
 }