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：
有n道题，每道题做对得一分，做错不得分，有一个人说“第一个人得X分，第二个人得Y分”，有这两个人关于这n道题得答案，判断一下这句话能否正确。
思路：
这答案中会有num数目得答案两个人是完全相同得，至于这num道题究竟对几道我们不需要考虑，（x-y）得绝对值得到的是两个人做对的题目得差，这个值肯定要小于（n-num）这个值，还要满足条件就是说两个人除去共同作对的题目，其余作对的题目的和也是要小于等于（n-num）这个值。
#include<iostream>
#include<stdio.h>
#include<string>
#include<cmath>
using namespace std;
const int N=80009;
int n,num,score1,score2;
char ch1[N];
char ch2[N];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
num=0;
scanf("%d%d%d",&n,&score1,&score2);
scanf("%s",ch1);
scanf("%s",ch2);
int flag=0;
for(int i=0; i<n; i++)
{
if(ch1[i]==ch2[i])
num++;
}
int Min=min(score1,score2);
Min=min(num,Min);
if((abs(score1-score2)<=n-num)&&(score1+score2-2*Min<=n-num))
printf("Not lying\n");
else
printf("Lying\n");
}
return 0;
}
