[JZOJ4804 survey results] [simulation] [double] Pointer

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Subject to the effect:

Topic links: https://jzoj.net/senior/#main/show/4804

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Ideas:

Obviously, for the arbitrary determination $r$ , to meet the requirements of the subject is certainly left sequence is a contiguous segment $[l_{r1}\sim_{r2}]$ .
And this range is clearly monotone nature.
So you can use three pointers maintenance $l1,l2,r$ . Every time the $r$ pointer right one when maintenance $l1$ and $l2$ , then the answer to accumulate it.
Amortized time complexity is $O (n)$

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Code:

**#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;

const int N=200010;
int n,m,l1,l2,r,sum,a[N],L[N],R[N],cnt1[N],cnt2[N];
ll ans;

int main()
{
	freopen("survey.in","r",stdin);
	freopen("survey.out","w",stdout);
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for (int i=1;i<=m;i++)
	{
		scanf("%d%d",&L[i],&R[i]);
		if (!L[i]) sum++;
	}
	l1=l2=1;
	for (r=1;r<=n;r++)
	{
		cnt1[a[r]]++;
		cnt2[a[r]]++;
		if (cnt2[a[r]]==L[a[r]]) sum++;
		while (cnt1[a[r]]>R[a[r]])
		{
			cnt1[a[l1]]--;
			l1++;
		}
		while (sum==m&&l2<=r)
		{
			cnt2[a[l2]]--;
			if (cnt2[a[l2]]==L[a[l2]]-1) sum--;
			l2++;
		}
		ans+=max((ll)l2-(ll)l1,0LL);
	}
	printf("%lld\n",ans);
	return 0;
}