2017 Multi-University Training Contest

原题

Is Derek lying?

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

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，每个题有ABC三个选项，有一个答案是正确的。每道题做对得一分，做错不得分，给定命题：“第一个人得X分，第二个人得Y分”，然后有这两个人关于这n道题得答案，判断一下这个答案序列能否成立。

涉及知识及算法

考虑到只有两种情况下这是不正确的：

1.x+y超过了给定答案能够提供的最大分数，这是x+y的上界。注意，x+y是没有下界的（可以答对0道）

故x+y<=n+cnt

2.x与y的差值过大。例如，答案全为相同的，x与y的差值为1。处理这种情况时，找到差值的上界（差值只能从不相同的答案得到），比较即可。

故abs(x-y)<=n-cnt

代码

#include <iostream>#include <cstdio>#include <cmath>using namespace std;char s1[80005],s2[80005];int main(){    int a,b,n,t,cnt;    scanf("%d",&t);    while(t--)    {        scanf("%d%d%d",&n,&a,&b);        scanf("%s",s1);        scanf("%s",s2);        cnt=0;        for(int i=0;i<n;i++)        {            if(s1[i]==s2[i])            {                cnt++;   //相同个数            }        }        if(abs(a-b)<=n-cnt&&a+b<=n+cnt)        {            printf("Not lying\n");        }        else        {            printf("Lying\n");        }    }    return 0;}

文章部分内容转载自博客园博主Pic，链接http://www.cnblogs.com/liuzhanshan/p/7249358.html