2015
05-23

# String change

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.

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.

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.

3
ab
ba

bac
ddb

aaabb
cbccd

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.


3个及3个以上一定可以T^T

2个就暴力判一下

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 66;
char a[N], b[N];
bool check() {
int n = 60;
while(n -- > 0) {
swap(a[0], a[1]);
a[0] ++;
a[1] ++;
if(a[0] > 'z') a[0] = 'a';
if(a[1] > 'z') a[1] = 'a';

if(a[0] == b[0] && a[1] == b[1]) return true;
}
return false;
}

int main() {
int T, cas = 0;
scanf("%d", &T);
while(T-- > 0) {
scanf("%s%s", a, b);
int n = strlen(a);
bool ok = 0;
if(n == 2) {
if(check()) ok = 1;
else ok = 0;
} else {
int s1 = 0, s2 = 0;
for(int i = 0; i < n; i ++) {
s1 += a[i] - 'a';
s2 += b[i] - 'a';
}
if((s1+s2)&1) ok = 0;
else ok = 1;
}
if(ok) printf("Case #%d: YES\n", ++cas);
else printf("Case #%d: NO\n", ++cas);
}
return 0;
}