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1.[模板]前缀和

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#include <iostream>
#include <vector>
using namespace std;int main() {int n = 0, q = 0;cin >> n >> q;vector<int> arr(n+1);//开辟一个n+1的数组for(int i = 1; i <= n; i++) cin >> arr[i];//创建一个前缀和数组。vector的构造会自己初始化vector<long long> dp(n+1);//更新前缀和数组for(int i = 1; i<=n; i++) dp[i] = dp[i-1] + arr[i];//直接使用前缀和数组进行返回即可int l = 0, r = 0;while(q--){cin >> l >> r;cout << dp[r] - dp[l-1] << endl;//直接输出结果即可}return 0;
}

2.【模板】二维前缀和

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3.寻找数组的中心下标

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class Solution {
public:int pivotIndex(vector<int>& nums) {int n = nums.size();vector<int> f(n), g(n);//1.分别求出前缀和、后缀和数组for(int i = 1; i<n; i++)f[i] = f[i-1] + nums[i-1];for(int i = n-2; i>=0; i--)g[i] = g[i+1] + nums[i+1];//2.使用前缀和、后缀和数组for(int i = 0; i<n; i++)if(f[i] == g[i]) return i;return -1;}
};

4.除自身以外数组的乘积

链接: link

class Solution {
public:vector<int> productExceptSelf(vector<int>& nums) {int n = nums.size();vector<int> f(n), g(n);//1.先求出f和g数组f[0] = 1;//注意：细节问题一定要处理g[n-1] = 1;for(int i = 1; i<n; i++)f[i] = f[i-1] * nums[i-1];for(int i = n-2; i>=0; i--)g[i] = g[i+1] * nums[i+1];//2.使用两数组vector<int> ret(n);for(int i = 0; i<n; i++)ret[i] = f[i] * g[i];return ret;}
};

5.和为k的子数组

链接: link

class Solution {
public:int subarraySum(vector<int>& nums, int k) {unordered_map<int, int> hash;hash[0] = 1;int sum = 0, ret = 0;for(auto e : nums){sum += e;//计算当前位置的前缀和if(hash.count(sum - k)) ret += hash[sum-k];hash[sum]++;}return ret;}
};

6.和可被k整除的子数组

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class Solution {
public:int subarraysDivByK(vector<int>& nums, int k) {unordered_map<int, int> hash;hash[0] = 1;//细节问题：如果nums的和可被k整除，那么也要将次数+1int sum = 0, ret = 0;for(auto e : nums){sum += e;int r = (sum%k + k) % k;//求余数的方法if(hash.count(r)) ret += hash[r];//如果sum%k = 前缀和%k，那么就可以被k整除hash[r]++;}return ret;}
};

7.连续数组

链接: link

class Solution {
public:int findMaxLength(vector<int>& nums) {unordered_map<int, int> hash;hash[0] = -1;int sum = 0, ret = 0;for(int i = 0; i<nums.size(); i++){sum += nums[i] == 0 ? -1 : 1;//我们不需要将数组的0改为1，只需要在加的这个部分加-1就行了if(hash.count(sum)) ret = max(ret, i-hash[sum]);else hash[sum] = i;//此时存储的应该是下标}return ret;}
};

8.矩阵区域和

链接: link

class Solution {
public:vector<vector<int>> matrixBlockSum(vector<vector<int>>& mat, int k) {int m = 0, n = 0;m = mat.size();n = mat[0].size();//先计算出前缀和数组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]-dp[i-1][j-1]+mat[i-1][j-1];//前缀和数组的使用vector<vector<int>> ret(m, vector<int>(n));for(int i = 0; i<m; i++){for(int j = 0; j<n; j++){int x1 = 0, y1 = 0, x2 = 0, y2 = 0;x1 = max(0, i-k) + 1;y1 = max(0, j-k) + 1;x2 = min(m-1, i+k) + 1;y2 = min(n-1, j+k) + 1;ret[i][j] = dp[x2][y2] - dp[x1-1][y2] - dp[x2][y1-1] + dp[x1-1][y1-1];}}return ret;}
};