【多校训练】hdu 6045 Is Derek lying?

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
,s2
;

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");
}
}