HDU6794 Tokitsukaze and Multiple（前缀和/贪心/尺取法）

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 (a_i)(1≤(a_i)≤105) is the i-th element ofa

.

It is guaranteed that the sum of n in all test cases is no larger than (10^6)

.

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

啊我好菜我好菜我好菜Q^Q

放一道和这个题很类似的https://codeforces.com/problemset/problem/1333/C

题意就是求最大的不相交的和为p的倍数的连续子段的数目。

注意到这样一个性质：对这个数列求前缀和再取模，如果有两个位置的数相等，说明这段区间的和是p的倍数。因此可以求出当前位置取模过后的前缀和的值，如果这个值在map里出现过，就更新ans++，同时清空map并把当前前缀和的值设置为0（因为当前点到之前的点之间有线段覆盖的话，后面的点和之前的点之间就不能有线段了）再添加map[0]=1（别忘了取模后为0的情况）；没有出现过的话就把当前取模过后的前缀和的值添加到map里。

貌似还有DP做法：

last[i]数组记录最后i出现的位置 dp[i]=max(dp[i-1],dp[last[cur]]+1);

#include <bits/stdc++.h>
using namespace std;
int n, p, ans;
int sum[100005] = {0};
int main()
{
	int t;
	cin >> t;
	while(t--)
	{
		ans = 0;
		map<int, int> mp;
		scanf("%d%d", &n, &p);
		mp[0] = 1;
		for(int i = 1; i <= n; i++)
		{
			int temp;
			scanf("%d", &temp);
			sum[i] = (sum[i - 1] + temp) % p;
			if(mp.count(sum[i]))
			{
				ans++;
				mp.clear();
				sum[i] = 0;
				mp[0] = 1;
			}
			else
			{
				mp[sum[i]] = 1;
			}
		}
		cout << ans << endl;
	}
}