hdu 4357 String change(推理题，4级)

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

Hint

Author

miketc@UESTC_Goldfinger

Source

​​2012 Multi-University Training Contest 6​​

Recommend

zhuyuanchen520

思路：推理后发现超过两张牌的情况，只要是奇偶一样都可以用变换。

#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\n");
        else printf("NO\n");
      }
      else
      {
        for(int i=0;i<len;++i)
        {
          a^=s[i]&1;
          b^=t[i]&1;
        }
        if(a==b)printf("YES\n");
        else printf("NO\n");
      }
    }
  }
  return 0;
}