HDu 4357 String change(推理問題、4級)

String change
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 802    Accepted Submission(s): 370
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
 
Recommend
zhuyuanchen520
考え方:推理して2枚以上のカードを発見した場合、パリティのように変換することができます.

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int mm=99;
char s[mm],t[mm];
int cas;
int main()
{
  while(~scanf("%d",&cas))
  {
    for(int ca=1;ca<=cas;++ca)
    { scanf("%s%s",s,t);
      printf("Case #%d: ",ca);
      int len=strlen(s);
      bool a=0,b=0;
      bool flag=0;
      int k1,k2;
      if(len==2)
      { k1=(s[0]-t[0]+26)%26;
        k2=(s[1]-t[1]+26)%26;
        if(k1==k2&&k1%2==0)flag=1;
        k1=(s[0]-t[1]+26)%26;
        k2=(s[1]-t[0]+26)%26;
        if(k1==k2&&k1%2==1)flag=1;
        if(flag)printf("YES
");
        else printf("NO
");
      }
      else
      {
        for(int i=0;i<len;++i)
        {
          a^=s[i]&1;
          b^=t[i]&1;
        }
        if(a==b)printf("YES
");
        else printf("NO
");
      }
    }
  }
  return 0;
}