Codeforces648div2C

Codeforces648div2C
원제:
After the mysterious disappearance of Ashish, his two favourite disciples Ishika and Hriday, were each left with one half of a secret message. These messages can each be represented by a permutation of size n. Let’s call them a and b.
Note that a permutation of n elements is a sequence of numbers a1,a2,…,an, in which every number from 1 to n appears exactly once.
The message can be decoded by an arrangement of sequence a and b, such that the number of matching pairs of elements between them is maximum. A pair of elements ai and bj is said to match if:
i=j, that is, they are at the same index. ai=bj His two disciples are allowed to perform the following operation any number of times:
choose a number k and cyclically shift one of the permutations to the left or right k times. A single cyclic shift to the left on any permutation c is an operation that sets c1:=c2,c2:=c3,…,cn:=c1 simultaneously. Likewise, a single cyclic shift to the right on any permutation c is an operation that sets c1:=cn,c2:=c1,…,cn:=cn−1 simultaneously.
Help Ishika and Hriday find the maximum number of pairs of elements that match after performing the operation any (possibly zero) number of times.
Input The first line of the input contains a single integer n (1≤n≤2⋅105) — the size of the arrays.
The second line contains n integers a1, a2, …, an (1≤ai≤n) — the elements of the first permutation.
The third line contains n integers b1, b2, …, bn (1≤bi≤n) — the elements of the second permutation.
Output Print the maximum number of matching pairs of elements after performing the above operations some (possibly zero) times.
Examples
사고방식: 모든 원소가 a와 b의 위치에 있는 차이를 기록하고 마지막으로 가장 많이 반복되는 것을 선택한다.
ACcodes

#include 
using namespace std;
const int N = 2e5+10;
int visa[N]={0},visb[N]={0},e[N];
bool cmp(int a, int b)
{
	return a>b;
}
int main()
{
	int test=1;
	//scanf("%d",&test);
	while(test--) {
		int n;cin>>n;
		int a[N],b[N];
		
		for(int i=0;i<n;i++) 
			cin>>a[i],visa[a[i]]=i,e[i]=0;
		for(int i=0;i<n;i++) 
			cin>>b[i],visb[i]=visa[b[i]];
		int res=0;
		for(int i=0;i<n;i++)
		{
			res = max(res,++e[(visb[i]-i+n)%n]);
		}
		cout << res << endl;
	}
	return 0;
}