LG-P6006 [USACO20JAN]Farmer John Solves 3SUM G 题解

The number of triples in the interval (satisfying $toi+aj+ak=0$）

(The fourth question of the first test of GZEZ New Year's Day)

in the specified range $[l,r]$ ,$toi+aj+ak=$The number of triples of$0\; .$

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Ideas:

Preprocessing first, the time complexity is $O\left(n^2\right)$; query again, the time complexity is$O\left(1\right)$。

Preprocessing:

1. After determining the numbers at both ends, the middle number $=-a\left[i\right]-a\left[j\right]$ , and use buckets to count the number of each number, that is,$dp\left[i\right]\left[j\right]=cnt[m-a[i]-a[j\left]\right]$ （$dp$ ​​is used to record the items that meet the conditions in$i$ tojjThe number of triples of$j$$cnt\; is$ used as a bucket to count the number of occurrences of each number).

2. Remember to put cnt cnt outside the second level loop$cnt$ The used part is cleared.

3. dp[i][j] before $dp\left[i\right]\left[j\right]$ counts with$i$ headed,$j$ is the triple included at the end, if you want to get$[i,$For all triples in$j$$]\; ,\; use\; the\; following\; one-dimensional\; difference,\; that\; is:$

   $dp\left[i\right]\left[j\right]+=\left(dp\right[i+1\left]\right[j]+dp[i\left]\right[j-1]-dp[i+1\left]\right[j-1\left]\right)$

Inquire:

Enter $l,r$ , output$dp\left[l\right]\left[r\right]$ is enough.

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$Code$

#include<bits/stdc++.h>
using namespace std;

#define int long long 
int n, q;
const int m = 1000000;
const int maxn = 5005;
int a[maxn];
int dp[maxn][maxn];
int cnt[4000005];

signed main ()
{
    
    
	scanf ("%lld %lld", &n, &q);
	for (int i = 1; i <= n; i++)
	{
    
    
		scanf ("%lld", &a[i]);
	} 
	for (int i = 1; i <= n; i++)
	{
    
    
		for (int j = i + 1; j <= n; j++)
		{
    
    
			if (0 <= m - a[j] - a[i] and m - a[i] - a[j] <= 2 * m) dp[i][j] = cnt[m - a[i] - a[j]];
			cnt[a[j] + m]++;
		}
		for (int j = i + 1; j <= n; j++) cnt[a[j] + m]--;
	}
	for (int len = 3; len <= n; len++)
	for (int i = 1; i + len - 1 <= n; i++)
	{
    
    
		int j = i + len - 1;
		dp[i][j] += (dp[i + 1][j] + dp[i][j - 1] - dp[i + 1][j - 1]);
	}
	while (q--)
	{
    
    
		int l, r;
		scanf ("%lld %lld", &l, &r);
		printf ("%lld\n", dp[l][r]);
	}
	return 0;
}

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