题目链接：http://codeforces.com/contest/1119/problem/A
Ilya lives in a beautiful city of Chordalsk.
There are n houses on the street Ilya lives, they are numerated from 1 to n from left to right; the distance between every two neighboring houses is equal to 11 unit. The neighboring houses are 1 and 2, 2 and 3, ..., n−1 and n. The houses n and 1 are not neighboring.
The houses are colored in colors c1,c2,…,cn so that the ii-th house is colored in the color cici. Everyone knows that Chordalsk is not boring, so there are at least two houses colored in different colors.
Ilya wants to select two houses ii and jj so that 1≤i<j≤n, and they have different colors: ci≠cj. He will then walk from the house ii to the house jj the distance of (j−i)units.
Ilya loves long walks, so he wants to choose the houses so that the distance between them is the maximum possible.
Help Ilya, find this maximum possible distance.
Input
The first line contains a single integer nn (3≤n≤300000) — the number of cities on the street.
The second line contains nn integers c1,c2,…,cn (1≤ci≤n) — the colors of the houses.
It is guaranteed that there is at least one pair of indices i and j so that 1≤i<j≤n and ci≠cj.
Output
Print a single integer — the maximum possible distance Ilya can walk.
Examples
input
Copy
5 1 2 3 2 3
output
Copy
4
input
Copy
3 1 2 1
output
Copy
1
input
Copy
7 1 1 3 1 1 1 1
output
Copy
4
Note
In the first example the optimal way is to walk from the first house to the last one, where Ilya can walk the distance of 5−1=4 units.
In the second example the optimal way is to either walk from the first house to the second or from the second to the third. Both these ways have the distance of 1 unit.
In the third example the optimal way is to walk from the third house to the last one, where Ilya can walk the distance of 7−3=4 units.
颜色不同的最大距离要么以左端点为起点，要么以右端点为终点
为什么不可能中间某两点的距离最大？
假设中间两点距离最大 a______b
左边肯定是b,右边肯定是a ba______ba,假设不成立
#include<bits/stdc++.h>
using namespace std;
int n,A[300001],ans;
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++)scanf("%d",&A[i]);
for(int i=1;i<=n;i++){
if(A[i]!=A[1])ans=max(ans,i-1);
if(A[i]!=A[n])ans=max(ans,n-i);
}
printf("%d\n",ans);
return 0；
}