hdu6045 Is Derek lying? -逻辑题-2017多校联盟2 1题
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Is Derek lying?
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 648 Accepted Submission(s): 371
Problem Description
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 integersN ,X ,Y ,the meaning is mentioned above.
The second line consists ofN characters,each character is “A ” “B ” or “C ”,which represents the answer of Derek for each question.
The third line consists ofN 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
For each test case,there will be three lines.
The first line consists of three integers
The second line consists of
The third line consists of
Data Range:
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 ”.
Please print “
Sample Input
23 1 3AAAABC5 5 0ABCBCACBCB
Sample Output
Not lyingLying
Source
2017 Multi-University Training Contest - Team 2
/*题意:n个选择题,选项ABC,两人分别作答,并给出两人各自得分,问此得分是否正确题解设N个题目中两人相同的答案数为s,那么不相同的答案数为N-s其中相同的答案中得分的题目数为a,不得分的题目数为b,即s = a+b。其中不相同的答案中Derek答对的有m个,那么 Alfia答对的有n个,其中n<=N-s-m个,因为有可能两人都错;则 x = a+m; y <= a+N-s-m = a+N-(a+b)-m = N-b-m那么 x+y < a+N-bx+y-a < N-b <= Nx+y-(a+b)<=Nx+y-s<=N ①两个人的分数差就差在两人答案不同的题目上,容易得证,0<=两人的分数差<=不同题目数,即|x-y| <= N-s ②满足上述两个条件即可*/#include <iostream>#include <cstdio>#include <algorithm>#include <string.h>#include <stdlib.h>#include <set>#include <algorithm>using namespace std;int main(){ int t; int x,y; int n; scanf("%d",&t); while(t--){ scanf("%d%d%d",&n,&x,&y); char a[80010],b[80010]; scanf("%s%s",a,b); int s = 0; for(int i = 0;i<n;++i){ if(a[i]==b[i]) s++; } if(n-s >= abs(x-y) && x+y-s<=n){ printf("Not lying\n"); } else printf("Lying\n"); } return 0;}
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