HDOJ 6045-Is Derek lying?
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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
题目链接:点击打开链接
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
2
3 1 3
AAA
ABC
5 5 0
ABCBC
ACBCB
Sample Output
Not lying
Lying
2
3 1 3
AAA
ABC
5 5 0
ABCBC
ACBCB
Sample Output
Not lying
Lying
题意:D和A是好朋友,同时参加考试比赛,比赛完后,系统会告诉D他们俩的总得分情况,D比较顽皮,他有时不会向A汇报正确的得分情况,做对一题得一分,告诉你有多少道题,以及D告诉A他和A的总分,然后分别给出D和A的答案,问你D是否对A撒谎了。
分析:假设一个人的答案全部正确为n,c是他们一样答案的个数,那么他们的的总得分一定不大于n+c,他们得分差一定不大于n-c。
#include<cstdio>#include<cstring>#include<cmath>using namespace std;char s1[300010],s2[300010];int main(){ int t; scanf("%d",&t); while(t--) { int a,b,n; scanf("%d %d %d",&n,&a,&b); scanf("%s",s1); scanf("%s",s2); int c=0; for(int i=0;i<n;i++) { if(s1[i]==s2[i]) c++;//c是记录他们答案相同的个数 } if(a+b<=n+c&&abs(a-b)<=n-c) { printf("Not lying\n"); } else { printf("Lying\n"); } } return 0;}
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