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
题意：
给出n道题，A和B的分数，再给出两个人的选项。判断是否有这种可能。
思路：
两人不同的题数不能小于分差，并且如果一题不同，那两人这题的得分和最多为1，如果一题相同，得分和最多为2，总得分和必须不小于总分和。
//
// main.cpp
// 1001
//
// Created by zc on 2017/7/27.
// Copyright © 2017年 zc. All rights reserved.
//
#include <iostream>
#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
using namespace std;
const int N=110000;
char s1[N],s2[N];
int main(int argc, const char * argv[]) {
int T;
cin>>T;
while(T--)
{
int n,x,y;
scanf("%d%d%d",&n,&x,&y);
scanf("%s%s",s1,s2);
int dif=0;
for(int i=0;i<n;i++) if(s1[i]!=s2[i]) dif++;
if(dif<abs(x-y)||dif+(n-dif)*2<(x+y)) printf("Lying\n");
else printf("Not lying\n");
}
}
