Codeforces Gym-101808C（字符串）

C. Help Shahhoud

time limit per test

2.0 s

memory limit per test

256 MB

input

standard input

output

standard output

Shahhoud is participating in the first Div.3 contest on Codeforces, the first problem was:

Given two strings A and B of equal length N (N is odd), find the minimum number of steps needed to change A into B, or print  - 1 if it's impossible.

In each step, you can choose an odd integer x such that 1 ≤ x ≤ N, and reverse the substring of length x that is centered in the middle of A. For example, performing a step with x = 3 on the string "abcde" results in "adcbe" and applying x = 5 on "abcde" results in "edcba".

Can you help Shahhoud solve the problem?

Input

The first line contains one integer T, the number of test cases.

Each test case consists of two lines, the first contains the string A, and the second contains the string B. (1 ≤ |A| = |B| ≤ 10⁵) |A| = |B| is odd.

Both strings consist of lowercase English letters.

Output

For each test case, print one line containing one integer,  - 1 if A can't be changed into B, or the minimum number of steps to change Ainto B.

Example

input

Copy

1
abcxdef
fecxdba

output

Copy

2

此题也不难，但是要注意细节，下面第一个代码TLE了，原因是当输出-1的时候没用break及时退出，第二个代码AC。

代码一（TLE）：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
int main(){
    string str1,str2;
    int n;
    int j;
    int i;
    int len;
    int sum;
    cin>>n;
    while(n--){
        cin>>str1;
        cin>>str2;
        len=str1.length();
        sum=0;
        for( i=0,j=len-1;i<len;i++,j--){
            if((str1[i]!=str2[i])&&(str1[i]!=str2[j])) 
            printf("-1\n");
        }
        for(int i=0,j=len-1;i<(len-1)/2;i++,j--){
            if(sum%2==0){
                if(str1[i]!=str2[i])
                sum++;
            }
            else{
                if(str1[i]!=str2[j])
                sum++;
            }
        }
        cout<<sum<<endl;
}
    return 0;
}

代码二（AC）：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
int main(){
    string str1,str2;
    int n;
    int j;
    int i;
    int len;
    int sum;
    cin>>n;
    while(n--){
        cin>>str1;
        cin>>str2;
        len=str1.length();
        sum=0;
        for( i=0,j=len-1;i<len;i++,j--){
            if((str1[i]!=str2[i])&&(str1[i]!=str2[j])) 
            break;
        }
        if(i!=len) printf("-1\n");
        else{
        for(int i=0,j=len-1;i<(len-1)/2;i++,j--){
            if(sum%2==0){
                if(str1[i]!=str2[i])
                sum++;
            }
            else{
                if(str1[i]!=str2[j])
                sum++;
            }
        }
        cout<<sum<<endl;
    }
   }
    return 0;
}