Is Derek lying?
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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分析:这道题为简单的思维题,应该分几组情况进行讨论#include <iostream>#include <cstdio>#include <cstring>#include <string>using namespace std;const int mm = 8e4+100;char a[mm], b[mm];#define mem(a) memset(a, 0, sizeof(a))int main() {int t;scanf("%d",&t);while (t --) {mem(a), mem(b);int n, x, y;scanf("%d%d%d",&n,&x,&y);scanf("%s",a);scanf("%s",b);int num = 0;for (int i = 0; i<n; i++) {if (a[i] == b[i]) num ++ ;}int f = 0;if (x >= num && y >= num) {int x1 = x-num, y1 = y-num;int xx = x1+y1;if (xx > n-num) f = 1;}else if (x >= num && y < num) {int x1 = x-y;if (x1 > n-num) f = 1;}else if (x < num && y >= num) {int y1 = y-x;if (y1 > n-num) f = 1;}else if (x < num && y < num) {f = 0;}if (f == 0) printf("Not lying\n");else printf("Lying\n");}return 0;}
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