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
Sample Input 2 3 1 3 AAA ABC 5 5 0 ABCBC ACBCB Sample Output Not lying Lying
题意：
现在两个人同时回答一套题，ABC三个选项，只有一个是正确的，答对得一分，答错不得分，给出两个人的成绩和答案，判断这两个成绩是否合理。
比如一共五个题，两个人的得分都是五分，然后两个人的答案都不相同，那么表示不合理。
思路：
没什么技巧，就是推理题目。推理方法有很多种，只要合理就行。
思路一：记录两个人答案相同的个数，如果 两个人成绩的和小于等于题目总数加上相同的个数 并且 两个人的分数差小于等于题目总数减去相同的个数，则表示没说谎。
思路二：记录两人答案相同和不同的题数，如果 两个人成绩的和小于等于不同的个数加上两倍的相同的个数 并且 两个人分数差小于等于不相同的答案数，则表示没说谎
代码一：
/*
ACACBCA
BABACBB
*/
#include<stdio.h>
#include<math.h>
#include<cmath>
using namespace std;
char a1[80050],a2[80050];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n,x,y,i,j;
int butong=0,xiangtong=0;
scanf("%d%d%d",&n,&x,&y);
scanf("%s%s",a1,a2);
for(i=0;i<n;i++)//计算相同和不同的个数
{
if(a1[i]==a2[i])
xiangtong++;
else
butong++;
}
if(x+y<=n+xiangtong&&abs(x-y)<=n-xiangtong)
printf("Not lying\n");
else
printf("Lying\n");
}
}
代码二：
/*
ACACBCA
BABACBB
*/
#include<stdio.h>
#include<math.h>
#include<cmath>
using namespace std;
char a1[80050],a2[80050];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n,x,y,i,j;
int butong=0,xiangtong=0;
scanf("%d%d%d",&n,&x,&y);
scanf("%s%s",a1,a2);
for(i=0;i<n;i++)//计算相同和不同的个数
{
if(a1[i]==a2[i])
xiangtong++;
else
butong++;
}
if(abs(x-y)<=butong&&x+y<=butong+2*xiangtong)
printf("Not lying\n");
else
printf("Lying\n");
}
}