第十一屆河南ACM Gene mutation（模擬）

Ostrichcrab發表於2018-06-01
ACM

提交地址：點選開啟連結

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)
輸入
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)
輸出
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
樣例輸入
複製
161 8 5 7 9 1034 6 7
樣例輸出
複製
2
排序，再掃一遍陣列就行了
#include<bits/stdc++.h>
using namespace std;
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		int n1,n2,cnt=0;
		int a[20010],b[106],c[106];
		cin>>n1;
		for(int i = 0;i<n1;i++)
			cin>>a[i];
		cin>>n2;
		for(int i = 0;i<n2;i++)
			cin>>b[i];
		sort(b,b+n2);
		int tmp = b[0];
		for(int i = 0;i< n2;i++)
			b[i] = b[i]-tmp;
	//	for(int i = 0;i< n2;i++)
	//		cout<<b[i]<<" ";
		//	cout<<'\n';
		for(int i = 0;i<n1-n2+1;i++)
		{
			for(int j = 0,k=i;j<n2;j++)
			{
				c[j] = a[k++];
			}
			sort(c,c+n2);
			tmp = c[0];
			int flag = true;
			for(int j =0;j<n2;j++)
			{
				c[j] = c[j]- tmp; // cout<<c[j]<<' ';
				if(c[j]!=b[j]) {
					flag = false;
			
				}
			}
			if(flag) cnt++;
		}
	    cout<<cnt<<endl;
	}
	return 0;
}
/*
1
6
1 8 5 7 9 10
3
4 6 7
*/

            




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