找规律：String Change

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本文介绍了一个字符串转换问题，探讨了如何通过交换字符并增加其字母顺序值来将一个字符串转换成另一个字符串的方法。提供了具体的实现代码及样例输入输出，帮助读者理解解决这类问题的思路。

String change

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 370 Accepted Submission(s): 166

Problem Description

In this problem you will receive two strings S ₁ and S ₂ that contain only lowercase letters.
Each time you can swap any two characters of S ₁. After swap,both of the two letters will increase their value by one. If the previous letter is 'z',it will become 'a' after being swapped.
That is to say ,"a" becomes "b","b" becomes "c"....."z" becomes "a" and so on.
You can do the change operation in S ₁ as many times as you want.
Please tell us whether you can change S ₁ to S ₂ after some operations or not.

Input

There are several cases.The first line of the input is a single integer T (T <= 41) which is the number of test cases.Then comes the T test cases .

For each case,the first line is S ₁,the second line is S ₂.S ₁ has the same length as S ₂ and the length of the string is between 2 and 60.

Output

For each case,output "Case #X: " first, X is the case number starting from 1.If it is possible change S ₁ to S ₂ output "YES",otherwise output "NO".

Sample Input

  
   3
ab
ba

bac
ddb

aaabb
cbccd

Sample Output

  
   Case #1: NO
Case #2: YES
Case #3: YES

   
    
     Hint
    
For the first case,it's impossible to change "ab" to "ba" .

For the second case,swap(S1[0],S1[2])->swap(S1[1],S1[2]),meanwhile:bac->dac->ddb.

For the third case,swap(S1[0],S1[3])->swap(S1[1],S1[2])->swap(S1[2],S1[3])->swap(S1[3],S1[4]),
meanwhile:aaabb->caabb->cbbbb->cbccb->cbccd.

Author

miketc@UESTC_Goldfinger

Source

2012 Multi-University Training Contest 6

这个题完全是找规律= =。

当长度为2的时候特判一下，其他长度如果同奇偶就YES，否则NO

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>

using namespace std;

char str1[70], str2[70];
int num1[70], num2[70];

int main(){
    int t;
    int cnt = 0;
    scanf("%d", &t);
    while(t --){
        cnt ++;
        scanf("%s %s", str1, str2);
        int len = strlen(str1);
        for(int i = 0; i < len; i ++){
            num1[i] = str1[i] - 'a' + 1;
            num2[i] = str2[i] - 'a' + 1;
        }
        printf("Case #%d: ", cnt);
        if(len == 2){
            if(num1[0] - num2[0] == num1[1] - num2[1] && (num1[0] - num2[0]) % 2 == 0)
                printf("YES\n");
            else if(num1[0] - num2[1] == num1[1] - num2[0] && (int)abs((float)num1[0] - (float)num2[1]) % 2)
                printf("YES\n");
            else
                printf("NO\n");
        }
        else{
            int sum = 0;
            for(int i = 0; i < len; i ++){
                sum += num1[i] - num2[i];
            }
            if((int)abs((sum)) % 2 == 0)
                printf("YES\n");
            else
                printf("NO\n");
        }
    }
    return 0;
}