多校第三场——1004 Tokitsukaze and Multiple

Problem Description
Tokitsukaze has a sequence of length n, denoted by a.

Tokitsukaze can merge two consecutive elements of a as many times as she wants. After each operation, a new element that equals to the sum of the two old elements will replace them, and thus the length of a will be reduced by 1.

Tokitsukaze wants to know the maximum possible number of elements that are multiples of p she can get after doing some operations (or doing nothing) on the sequence a.

Input
There are several test cases.

The first line contains an integer T (1≤T≤20), denoting the number of test cases. Then follow all the test cases.

For each test case, the first line contains two integers n and p (1≤n,p≤105), denoting the length of the sequence and the special number, respectively.

The second line contains n integers, where the i-th integer ai (1≤ai≤105) is the i-th element of a.

It is guaranteed that the sum of n in all test cases is no larger than 106.

Output
For each test case, output in one line the maximum possible number of elements that are multiples of p after doing some operations.

Sample Input
2
5 3
2 1 3 2 1
3 1
123 456 789

Sample Output
3
3

题意：给你一个数字序列，问你最多能选出几个子段，使得每段和恰好是p的倍数。

思路：考虑前缀数组。从左至右判断，当有两个前缀和相等时，这段区间和就一定是 p 的倍数

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 5;
int T;
int n, p;
ll sum[N];
int num[N];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> T;
    while (T--)
    {
        memset(num, -1, sizeof num);
        cin >> n >> p;
        for (int i = 1; i <= n; i++)
        {
            ll x;
            cin >> x;
            sum[i] = sum[i - 1] + x;
        }
        int ed = 0;
        num[0] = 0;
        int ans = 0;
        for (int i = 1; i <= n; i++)
        {
            int temp = sum[i] % p;
            if (num[temp] >= ed)
            {
                ans++;
                ed = i;
            }
            num[temp] = i;
        }
        cout << ans << endl;
    }
}