我们先来实现一个二维数组的前缀：
int main()
{
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];
}
}
vector<vector<long long>> dp(n + 1, vector<long long>(m + 1));
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];
int x1 = 0, y1 = 0, x2 = 0, y2 = 0;
while (q > 0) {
q--;
cin >> x1 >> y1 >> x2 >> y2;
cout << dp[x2][y2] - dp[x1 - 1][y2] - dp[x2][y1 - 1] + dp[x1 - 1][y1 - 1] << endl;
}
}
}
return 0;
}
class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
int n=nums.size();
vector<int> f(n),g(n);
f[0]=1;
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];
}
vector<int> ret(n);
for(int i=0;i<n;i++){
ret[i]=f[i]*g[i];
}
return ret;
}
};
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;
if(hash.count(sum-k)) ret+=hash[sum-k];
hash[sum]++;
}
return ret;
}
};
class Solution {
public:
int pivotIndex(vector<int>& nums) {
int n=nums.size();
vector<int> f(n),g(n);
for(int i=1;i<n;i++){
f[i]=f[i-1]+nums[i-1];
}
for(int j=n-2;j>=0;j--){
g[j]=g[j+1]+nums[j+1];
}
for(int i=0;i<n;i++){
if(f[i]==g[i]){
return i;
}
}
return -1;
}
};
class Solution {
public:
int findMaxLength(vector<int>& nums) {
int n=nums.size();
unordered_map<int,int> hash;
hash[0]=-1;
int sum=0,ret=0;
for(int i=0;i<n;i++){
sum+=nums[i]==0?-1:1;
if(hash.count(sum)) ret=max(ret,i-hash[sum]);
else hash[sum]=i;
}
return ret;
}
};
class Solution {
public:
vector<vector<int>> matrixBlockSum(vector<vector<int>>& mat, int k) {
int m=mat.size(),n=mat[0].size();
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];
}
}
vector<vector<int>> ret(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;
ret[i][j]=dp[x2][y2]-dp[x1-1][y2]-dp[x2][y1-1]+dp[x1-1][y1-1];
}
}
return ret;
}
};