2018湘潭邀請賽G題題解

2018湘潭邀請賽G題題解

G. String Transformation

Bobo has a string S = s₁s₂ . . . s_n consists of letter a, b and c. He can transform the string by inserting or deleting substrings aa, bb and abab.

Formally, A = u ◦ w ◦ v (“◦” denotes string concatenation) can be transformed into A^r = u ◦ v and vice versa where u, v are (possibly empty) strings and w ∈ {aa, bb, abab}.

Given the target string T = t₁t₂ . . . t_m, determine if Bobo can transform the string S into T .

 

Input

The input consists of several test cases and is terminated by end-of-file.

The first line of each test case contains a string s₁s₂ . . . s_n. The second line contains a string t₁t₂ . . . t_m.

 

Output

For each test case, print Yes if Bobo can. Print No otherwise.

 

Constraint

• 1 n, m 104

• s₁, s₂, . . . , s_n, t₁, t₂, . . . , t_ma, b, c

• The sum of n and m does not exceed 250, 000.

 

Sample Input

ab ba ac ca a ab

 

Sample Output

Yes No No

 

Note

For the first sample, Bobo can transform as ab => aababb => babb => ba.

 

 

 

題意：

 

讀完題目，簡化後就是c的位置不能變化,c之間的a和b的位置可以變化。a和b位置可以交換的原因就是有abab的存在。

 

再簡化就是統計c之間（或者c至開頭或者c至結尾）a和b的數目。詳見代碼吧。

 

 

AC代碼：

 

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int SIZE = 1e4+5;
int main() {
    char arr1[SIZE], arr2[SIZE];
    while(scanf("%s", arr1) != EOF) {
        scanf("%s", arr2);
        int len1 = strlen(arr1), len2 = strlen(arr2);
        int a1a[SIZE] = {0};
        int a1b[SIZE] = {0};
        int a2a[SIZE] = {0};
        int a2b[SIZE] = {0};
        int x = 0, y = 0;
        for(int i = 0; i < len1; i++) {
            if(arr1[i] != 'c') {
                if(arr1[i] == 'a') a1a[x]++;
                else if(arr1[i] == 'b') a1b[x]++;
            } else {
                x++;
            }
        }
        for(int i = 0; i < len2; i++) {
            if(arr2[i] != 'c') {
                if(arr2[i] == 'a') a2a[y]++;
                else if(arr2[i] == 'b') a2b[y]++;
            } else {
                y++;
            }
        }
        bool flag = true;
        if(x != y) printf("No\n");
        else {
            for(int i = 0; i <= x; i++) {
                if(a1a[i] % 2 != a2a[i] % 2 || a1b[i] % 2 != a2b[i] % 2) {
                    printf("No\n");
                    flag = false;
                    break;
                }
            }
            if(flag) printf("Yes\n");
        }
    }
    return 0;
}