HDU 6045 (2017 多校训练赛2 1001）Is Derek lying?

2017 Multi-University Training Contest - Team 2 1001

Is Derek lying?

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

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 1point.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 Alfiawill 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

23 1 3AAAABC5 5 0ABCBCACBCB

 

Sample Output

Not lyingLying

 

题意：有n个问题，两个人分别的得分为 a和b，给出两个人每一题的答案，问是否有矛盾。
分析：只需要知道什么情况会出现矛盾就可以了设 sam为两个人相同的题目那么 n-sam为他们不相同的题目所以   他们的分数差必须小于他们不相同的题目的数量   abs(a-b)<=n-sam他们未得到的分之和必须大于题目不相同的题目的数量   n-sam<=n-a+n-b
AC代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;char aima[800000],aimb[800000];int abs(int a){    return a>0?a:-a;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        int n,a,b;        scanf("%d%d%d",&n,&a,&b);        int sam=0;        scanf("%s",aima);scanf("%s",aimb);        for(int i=0;i<n;i++)        if(aima[i]==aimb[i])        sam++;               //相同题目数量         int aa=abs(a-b);    //分数差距         if(aa<=n-sam&&n-sam<=n-a+n-b)        printf("Not lying\n");        else        printf("Lying\n");     }}