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[Template] Praefixio una dimensiva sum

[Template] Duo dimensiva praepositionis sum

Reperio a centro indicem ordinata

Product vestit quam sui

matrix area sum

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[Template] Praefixio una dimensiva sum

Si violente utimur solutione;Ordo omni tempore percurrendus est, numerus q, sictempus complexionemnimis alta est, hoc tempore summam praepositionem construimusSumma cuiusque intervalli intra 1-n intervallum reponitur , debes prima n supellex et directa accessus ad dp praepositione et subscriptio positio ordinata. ut infra ostende codice:

#include<iostream>
#include<vector>
using namespace std;
 
int main()
{
	// 读入数据
	int n, q; cin >> n >> q;
	// n + 1 添加了虚拟节点0
	vector<int> arr(n + 1); // 默认全部为0
	for (int i = 1; i <= n; i++)
		cin >> arr[i];
 
	// 预处理出一个前缀和数组
	vector<long long> dp(n + 1); // 防止溢出
	for (int i = 1; i <= n; i++)
		dp[i] = dp[i - 1] + arr[i];
 
	// 使用前缀和数组
	int l = 0, r = 0;
	while (q--)
	{
		cin >> l >> r;
		cout << dp[r] - dp[l - 1] << endl;
	}
	return 0;
}

---

[Template] Duo dimensiva praepositionis sum

preprocessingpraepositionem sum matrix quae voluntasSumma omnium elementorum ab (1, 1) ad positiones (i, j)exstat in hoc dp ordinata, perArea calculi modusAd finalem responsionem inveniendam, in codice talis est;

int main()
{
	// 读入数据
	int n, m, q; cin >> n >> m >> q;
	vector<vector<int>> arr(n + 1, vector<int>(m + 1));
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++)
			cin >> arr[i][j];
 
	// 预处理一个前缀和数组
	vector<vector<long long>> dp(n + 1, vector<long long>(m + 1)); // 防止溢出
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++)
			dp[i][j] = dp[i - 1][j] + dp[i][j - 1] + arr[i][j] - dp[i - 1][j - 1];
 
	// 使用前缀和数组
	while (q--)
	{
		int x1, y1, x2, y2; cin >> x1 >> y1 >> x2 >> y2;
		cout << dp[x2][y2] - dp[x2][y1 - 1] - dp[x1 - 1][y2] + dp[x1 - 1][y1 - 1] << endl;
	}
	return 0;
}

---

Reperio a centro indicem ordinata

Nota in extremis casibus hicNon opus est aperire n+1 spatium praepositionis et ordinata, eo quod in prima acie est utendum ut centrum subscript huius quaestionis.

class Solution {
public:
	int pivotIndex(vector<int>& nums) {
		int n = nums.size();
		vector<int> f(n), g(n);
		// 预处理前缀和数组  从左向右
		for (int i = 1; i < n; i++)
			f[i] = f[i - 1] + nums[i - 1];
		// 预处理后缀和数组  从右向左
		for (int i = n - 2; i >= 0; i--)
			g[i] = g[i + 1] + nums[i + 1];
		for (int i = 0; i < n; i++)
		{
			if (g[i] == f[i])
				return i;
		}
		return -1;
	}
};

---

Product vestit quam sui

Sensus similis praecedenti quaestioni est, sed notandum est f(0) et g(n-1) casuum finium initialem esse 1 loco 0. In codice talis est:

class Solution {
public:
	vector<int> productExceptSelf(vector<int>& nums) {
		int n = nums.size();
		vector<int> f(n), g(n), ret(n);
		// 处理边界情况
		f[0] = 1; g[n - 1] = 1;
		// 预处理前缀积数组  从左向右
		for (int i = 1; i < n; i++)
			f[i] = f[i - 1] * nums[i - 1];
		// 预处理后缀积数组  从右向左
		for (int i = n - 2; i >= 0; i--)
			g[i] = g[i + 1] * nums[i + 1];
		for (int i = 0; i < n; i++)
			ret[i] = f[i] * g[i];
		return ret;
	}
};

---

matrix area sum

Nota: Praepositiones duae dimensiva et ordinata unum ordinem et unam columnam habere debent, alioquin extra limites accessus occurret. Praeterea subscripta necesse est aptari inter dp ordinatam et ans ordinatas ad positiones adaequare. ans[ 0 ][ 0 ] huic loco respondet dp [ 1 ][ 1]. ut infra ostende codice:

class Solution {
public:
	vector<vector<int>> matrixBlockSum(vector<vector<int>>& mat, int k) {
		int m = mat.size(), n = mat[0].size();  // m 为行 n 为列
		// 预处理一个二维前缀和数组 dp
		vector<vector<int>> dp(m + 1, vector<int>(n + 1));
		for (int i = 1; i <= m; i++)
			for (int j = 1; j <= n; j++)
				dp[i][j] = dp[i - 1][j] + dp[i][j - 1] + mat[i - 1][j - 1] - dp[i - 1][j - 1];
		// 存放答案的二维数组 ans
		vector<vector<int>> ans(m, vector<int>(n));
		for (int i = 0; i < m; i++)
		{
			for (int j = 0; j < n; j++)
			{
				int x1 = max(0, i - k) + 1, y1 = max(0, j - k) + 1;
				int x2 = min(m - 1, i + k) + 1, y2 = min(n - 1, j + k) + 1;
				ans[i][j] = dp[x2][y2] - dp[x1 - 1][y2] - dp[x2][y1 - 1] + dp[x1 - 1][y1 - 1];
			}
		}
		return ans;
	}
};