Problem Statement |
| | We have a string originalWord. Each character of originalWord is either 'a' or 'b'. Timmy claims that he can convert it to finalWord using exactly k moves. In each move, he can either change a single 'a' to a 'b', or change a single 'b' to an 'a'.
You are given the strings originalWord and finalWord, and the int k. Determine whether Timmy may be telling the truth. If there is a possible sequence of exactly k moves that will turn originalWord into finalWord, return "POSSIBLE" (quotes for clarity). Otherwise, return "IMPOSSIBLE". |
Definition |
| | | Class: | EasyConversionMachine | | Method: | isItPossible | | Parameters: | string, string, int | | Returns: | string | | Method signature: | string isItPossible(string originalWord, string finalWord, int k) | | (be sure your method is public) | |
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Notes |
| - | Timmy may change the same letter multiple times. Each time counts as a different move. |
Constraints |
| - | originalWord will contain between 1 and 50 characters, inclusive. |
| - | finalWord and originalWord will contain the same number of characters. |
| - | Each character in originalWord and finalWord will be 'a' or 'b'. |
| - | k will be between 1 and 100, inclusive. |
Examples |
| 0) |
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| | | | Returns: "IMPOSSIBLE" | | It is not possible to reach finalWord in fewer than 4 moves. | | |
| 1) |
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| | | | Returns: "IMPOSSIBLE" | | The number of moves must be exactly k=1. | | |
| 2) |
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| | | | Returns: "POSSIBLE" | | Use each move to change each of the letters once. | | |
| 3) |
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| | | | Returns: "POSSIBLE" | The following sequence of 4 moves does the job:
aaa -> baa -> bab -> aab -> bab | | |
| 4) |
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| | "aababbabaa" | "abbbbaabab" | 9 | | Returns: "IMPOSSIBLE" | | | |
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字符串水題！
249.23PT！！！
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <cassert>
#include <iostream>
#include <sstream>
#include <fstream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <algorithm>
#define min(x,y) (x<y?x:y)
#define max(x,y) (x>y?x:y)
#define swap(t,x,y) (t=x,x=y,y=t)
#define clr(list) memset(list,0,sizeof(list))
using namespace std;
class EasyConversionMachine{
public:
string isItPossible(string originalWord, string finalWord, int k)
{
int n=originalWord.size();
int tot=0;
for (int i=0;i<n;i++)
if (originalWord[i]!=finalWord[i])
tot++;
if (tot<=k && (k-tot)%2==0) //傻叉，這里剛剛開始搞反了，testing WA了一次
return "POSSIBLE";
return "IMPOSSIBLE";
}
};