jzoj4307. 【NOIP2015模拟11.3晚】喝喝喝（枚举）

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4307. 【NOIP2015模拟11.3晚】喝喝喝

Description

Input

Output

Sample Input
3 2
5 3 1

Sample Output
4

Data Constraint

分析：枚举每个区间终点，通过枚举倍数判定区间起点。

代码

#include <cstdio>
#define N 100005
#define ll long long
using namespace std;

int n,k,a[N];
int f[N];
ll ans;

int main()
{
	freopen("drink.in","r",stdin);
	freopen("drink.out","w",stdout);
	scanf("%d%d", &n, &k);
	int mx = 0;
	for (int i = 1; i <= n; i++)
	{
		scanf("%d", &a[i]);
		if (a[i] > mx) mx = a[i];
	}
	int l = 1, r = 2;
	f[a[l]]++;
	while (r <= n)
	{
		int i = 0;
		int s = a[r] * i + k;
		if (a[r] > k)
		while(s <= mx)
		{
			while (f[s]) f[a[l++]]--;
			i++;
			s = a[r] * i + k;
		}
		ans += r - l;
		f[a[r++]]++;
	}
	printf("%lld", ans + n);
	fclose(stdin);
	fclose(stdout);
}