A. Omkar and Password

トランスポートドアA.Omkar and Password time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output
Lord Omkar has permitted you to enter the Holy Church of Omkar! To test your worthiness, Omkar gives you a password which you must interpret! A password is an array a of n positive integers. You apply the following operation to the array: pick any two adjacent numbers that are not equal to each other and replace them with their sum. Formally, choose an index i such that 1≤i For example, for array [7,4,3,7] you can choose i=2 and the array will become [7,4+3,7]=[7,7,7] Note that in this array you can’t apply this operation anymore. Notice that one operation will decrease the size of the password by 1 What is the shortest possible length of the password after some number (possibly 0) of operations?
Input Each test contains multiple test cases. The first line contains the number of test cases t(1≤t≤100). Description of the test cases follows.The first line of each test case contains an integer n(1≤n≤2⋅105) — the length of the password.The second line of each test case contains n integers a1,a2,…,an (1≤ai≤109) — the initial contents of your password. The sum of n over all test cases will not exceed 2⋅105
Output For each password, print one integer: the shortest possible length of the password after some number of operations.
Example Input

2
4
2 1 3 1
2
420 420

Output

1
2

Note In the first test case, you can do the following to achieve a length of 1 Pick i=2to get [2,4,1] Pick i=1to get [6,1] Pick i=1to get [7] In the second test case, you can’t perform any operations because there is no valid i that satisfies the requirements mentioned above.
題意:nサイズの配列をあげて、毎回2つの異なる数を合併することができて、最後に配列を圧縮できる最小の長さはいくらですか?答えは1かnしかありません.
考え方:この问题の考え方は简単で、答えは1あるいはnだけです.入力した配列の中に1つの数字が异なる限り出力1で、入力した数字はすべて同じでnです.
ACコード

#include 
using namespace std;
int main()
{
     
	int t,n,a[200005],flag;
	cin>>t;
	while(t--)
	{
     
		flag=0;
		cin>>n;
		cin>>a[1];
		for(int i=2;i<=n;i++)
		{
     
			cin>>a[i];
			if(a[i]!=a[i-1]) 
				flag=1;
		}
		if(flag==1)
			cout<<"1"<<endl;
		else
			cout<<n<<endl;
	} return 0;
}