思路：
- 现有一个长度为n的数组，求出将该数组分割成任意长度的区间和为p的倍数的区间数max。
- 用前缀和记录下，遇到为p的倍数就 ans ++，并用vis数组标记下该sum[i]%p是可行答案。
代码实现：
#include<bits/stdc++.h>
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair<int, int>
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
const int inf = 0x7fffffff;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll mod = 1e9 + 7;
const int N = 2e5 + 5;
int t, n, p;
int a[N], sum[N], vis[N];
signed main()
{
IOS;
cin >> t;
while(t --){
cin >> n >> p;
for(int i = 1; i <= n; i ++){
cin >> a[i];
a[i] %= p;
sum[i] = sum[i-1] + a[i];
sum[i] %= p;
}
for(int i = 1; i <= p; i ++) vis[i] = -1;
int ans = 0, pos = 0; vis[0] = 0;
for(int i = 1; i <= n; i ++){
if(vis[sum[i]] >= pos){
ans ++;
pos =i;
}
vis[sum[i]] = i;
}
cout << ans << endl;
}
return 0;
}