河南省第十一届ACM Gene mutation

不念不忘少年蓝@ 2023-06-13 11:20 1阅读 0赞

题目描述  
Gene mutation is the sudden and inheritable mutation of genomic DNA molecules. From the molecular level, gene mutation refers to the change of the composition or sequence of base pairs in the structure of a gene. Although the gene is very stable, it can reproduce itself accurately when the cell divides. Under certain conditions, the gene can also suddenly change from its original existence to another new form of existence.  
A genome sequence might provide answers to major questions about the biology and evolutionary history of an organism. A 2010 study found a gene sequence in the skin of cuttlefish similar to those in the eye’s retina. If the gene matches, it can be used to treat certain diseases of the eye.  
A gene sequence in the skin of cuttlefish is specified by a sequence of distinct integers (Y1,Y2, …Yc). it may be mutated. Even if these integers are transposed ( increased or decreased by a common amount ) , or re-ordered , it is still a gene sequence of cuttlefish. For example, if “4 6 7” is a gene sequence of cuttlefish, then “3 5 6” (-1), “6 8 9” ( +2), “6 4 7” (re-ordered), and “5 3 6” (transposed and re-ordered) are also ruminant a gene sequence of cuttlefish.  
Your task is to determine that there are several matching points at most in a gene sequence of the eye’s retina (X1,X2, …, Xn)  
输入  
The first line of the input contains one integer T, which is the number of test cases (1<=T<=5). Each test case specifies:

*  Line 1: n ( 1 ≤ n ≤ 20,000 )
 *  Line 2: X1 X2… Xn ( 1 ≤ Xi ≤ 100 i=1…. n)
 *  Line 3: c ( 1 ≤ c≤ 10 )
 *  Line 4: Y1 Y2… Yc ( 1 ≤ Yi ≤ 100 i=1…. c)  
    输出  
    For each test case generate a single line: a single integer that there are several matching points. The matching gene sequence can be partially overlapped  
    样例输入 Copy  
    1  
    6  
    1 8 5 7 9 10  
    3  
    4 6 7  
    样例输出 Copy  
    2

package ACM2018;
    
    import java.util.Arrays;
    import java.util.Scanner;
    
    public class F { 
    	public static int T,n,m;
    	public static int bigsz[];
    	public static int smlsz[];
    	public static int celsz[];
    	public static int ans[];
    	public static void main(String[] args) { 
    		Scanner sc=new Scanner(System.in);
    		T=sc.nextInt();
    		ans=new int [T];
    		sc.nextLine();
    		for(int i=0;i<T;i++)
    		{ 
    			n=sc.nextInt();
    			bigsz=new int[n];
    			sc.nextLine();
    			for(int j=0;j<n;j++)
    			{ 
    				bigsz[j]=sc.nextInt();
    			}
    			sc.nextLine();
    			m=sc.nextInt();
    			smlsz=new int[m];
    			celsz=new int[m];
    			sc.nextLine();
    			for(int j=0;j<m;j++)
    			{ 
    				smlsz[j]=sc.nextInt();
    			}
    			Arrays.sort(smlsz);
    			ans[i]=check();
    			sc.nextLine();
    		}
    		for(int i=0;i<T;i++)
    		{ 
    			System.out.println(ans[i]);
    		}
    	}
    	public static int check()
    	{ 
    		int count=0;
    		for(int i=0;i<=n-m;i++)
    		{ 
    			boolean is=true;
    			for(int j=0;j<m;j++)
    			{ 
    				celsz[j]=bigsz[i+j];
    			}
    			Arrays.sort(celsz);
    			for(int j=0;j<m-1;j++)
    			{ 
    				if(celsz[j]-smlsz[j]==celsz[j+1]-smlsz[j+1])
    				{ 
    					continue;
    				}else { 
    					is=false;
    					break;
    				}
    			}
    			if(is)count++;
    		}
    		return count;
    	}
    }