找规律：String Change

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 S₁ and S₂ that contain only lowercase letters.
Each time you can swap any two characters of S₁. 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 S₁ as many times as you want.
Please tell us whether you can change S₁ to S₂ 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 S₁,the second line is S₂.S₁ has the same length as S₂ 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 S₁ to S₂ output "YES",otherwise output "NO".

 

Sample Input

3abbabacddbaaabbcbccd

 

Sample Output

Case #1: NOCase #2: YESCase #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;}