HDU 6045 Is Derek lying?

题目网址：http://acm.hdu.edu.cn/showproblem.php?pid=6045

题目：

Is Derek lying?

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 104    Accepted Submission(s): 65

Problem Description

Derek and Alfia are good friends.Derek is Chinese,and Alfia is Austrian.This summer holiday,they both participate in the summer camp of Borussia Dortmund.During the summer camp,there will be fan tests at intervals.The test consists of N choice questions and each question is followed by three choices marked “A” “B” and “C”.Each question has only one correct answer and each question is worth 1 point.It means that if your answer for this question is right,you can get 1 point.The total score of a person is the sum of marks for all questions.When the test is over,the computer will tell Derek the total score of him and Alfia.Then Alfia will ask Derek the total score of her and he will tell her: “My total score is X,your total score is Y.”But Derek is naughty,sometimes he may lie to her. Here give you the answer that Derek and Alfia made,you should judge whether Derek is lying.If there exists a set of standard answer satisfy the total score that Derek said,you can consider he is not lying,otherwise he is lying.

 

 

Input

The first line consists of an integer T,represents the number of test cases.

For each test case,there will be three lines.

The first line consists of three integers N,X,Y,the meaning is mentioned above.

The second line consists of N characters,each character is “A” “B” or “C”,which represents the answer of Derek for each question.

The third line consists of N characters,the same form as the second line,which represents the answer of Alfia for each question.

Data Range:1≤N≤80000,0≤X,Y≤N,∑Ti=1N≤300000

 

Output

For each test case,the output will be only a line.

Please print “Lying” if you can make sure that Derek is lying,otherwise please print “Not lying”.

 

Sample Input

2

3 1 3

AAA

ABC

5 5 0

ABCBC

ACBCB

 

Sample Output

Not lying

Lying

思路：

纯粹的思维题，用cnt记录两人答案不一样的题数。通过观察可以得到两个结论：

1.fabs(x-y)<=cnt;//两者答对题目之差要小于等于不相同的答案数，取等的条件是对于不相同的答案 都是同一人答对

2.x+y<=2*n-cnt//两者答对题目之和要小于等于 题目总数的两倍减去不相同的答案，取等的条件是两人相同答案的题 全对

代码：

 #include <cstdio>
 #include <cmath>
 const int N=;
 int main(){
     int t;
     int n,x,y;
     char D[N],A[N];
     scanf("%d",&t);
     while(t--){
         int cnt=;
         scanf("%d%d%d ",&n,&x,&y);
         gets(D);
         gets(A);
         for(int i=;i<n;i++){
             if(D[i]!=A[i])  cnt++;
         }
         if(fabs(x-y)>cnt || (x+y)>(*n-cnt))   printf("Lying\n");
         else printf("Not lying\n");
     }
     return ;
 }