2017 Multi-University Training Contest

题目：

Is Derek lying?

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

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 ofN 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 tellDerek the total score of him and Alfia.ThenAlfia will ask Derek the total score of her and he will tell her: “My total score is X,your total score is Y.”ButDerek 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 integerT,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

23 1 3AAAABC5 5 0ABCBCACBCB

 

Sample Output

Not lyingLying

题意：有n道题目，每道题目3个选择A、B、C，答对得一分。给你两个人的分数X和Y，判断是否有标准答案满足X和Y

思路：找到两个人不同的选项，如果不同的选项数目大于等于他们分数差距而且两个人分数和减去相同选项的数目小鱼等于n，说明没有说谎。

CODE:

#include<bits/stdc++.h>using namespace std;char a[80005],b[80005];int main(){        int t,n,x,y;        scanf("%d",&t);        while(t--){                scanf("%d%d%d%s%s",&n,&x,&y,a,b);                int num=0;                for(int i=0;i<n;i++) if(a[i]!=b[i]) num++;                if(abs(x-y)<=num&&x+y-(n-num)<=n) puts("Not lying");                else puts("Lying");        }    return 0;}