1.诸葛亮的排列组合
看成正方形的四个角，旋转三次，直接暴力枚举四种状态，一一对应
import java.util.Scanner;
public class A1 {
public static void main(String[] args) {
int a[][]=new int [5][5];
int b[][]=new int [5][5];
Scanner sc=new Scanner(System.in);
int t,k=1;
t=sc.nextInt();
while(k<=t){
for (int k2 = 1; k2 <= 2; k2++) {
for (int l = 1; l <=2; l++) {
a[k2][l]=sc.nextInt();
}
}
for (int k2 = 1; k2 <= 2; k2++) {
for (int l = 1; l <=2; l++) {
b[k2][l]=sc.nextInt();
}
}
int flag=0;
if(a[1][1]==b[1][1]&&a[1][2]==b[1][2]&&a[2][1]==b[2][1]&&a[2][2]==b[2][2])flag=1;
else if(a[1][1]==b[1][2]&&a[1][2]==b[2][2]&&a[2][1]==b[1][1]&&a[2][2]==b[2][1])flag=1;
else if(a[1][1]==b[2][2]&&a[1][2]==b[2][1]&&a[2][1]==b[1][2]&&a[2][2]==b[1][1])flag=1;
else if(a[1][1]==b[2][1]&&a[1][2]==b[1][1]&&a[2][1]==b[2][2]&&a[2][2]==b[1][2])flag=1;
if(flag==1){
System.out.println("Case #"+k+": "+"POSSIBLE");
}
else{
System.out.println("Case #"+k+": "+"IMPOSSIBLE");
}
k++;
}
}
}
2.华佗的药瓶，水题
import java.util.Scanner;
public class B {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
int k=0;
while(++k<=t) {
int n=sc.nextInt();
int m=2*n-1;
System.out.println("Case #"+k+": "+m);
}
}
}
3.水题
两个数组的对照，下标找对就行了
import java.util.Scanner;
public class C {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
int k=0;
while(++k<=t) {
int n=sc.nextInt();
int m=sc.nextInt();
int p=sc.nextInt();
int a[]=new int [n];
int b[]=new int [m];
for (int i = 0; i < a.length; i++)
a[i]=sc.nextInt();
for (int i = 0; i < b.length; i++)
b[i]=sc.nextInt();
int q=n-(m-1)*p;
int s1=0;
for (int i = 0; i < q; i++) {
int num=0;
for (int j = 0; j < b.length ; j++) {
//是否对应
if(b[j]==a[i+j*p] && i+j*p<a.length)
num++;
}
//如果全部对应
if(num==b.length) {
s1++;}
}
System.out.println("Case #"+k+": "+s1);
}
}
}