Enter an integer matrix of n rows and m columns, then enter a query q, each challenge comprising four integers x1, y1, x2, y2, upper-left coordinates and lower right coordinates represents a sub-matrices.
For each sub-query output matrix and all the numbers.
Input Format
The first line contains three integers n, m, q.
Next n lines, each line contains integers m represents an integer matrix.
Next q rows, each row comprising four integers x1, y1, x2, y2, represents a group of query.
Output Format
A total of q rows, each row outputs a result of the inquiry.
data range
. 1 ≤ n- , m ≤ 1000
. 1 ≤ Q ≤ 200000 ,
. 1 ≤ X . 1 ≤ X 2 ≤ n- ,
. 1 ≤ Y . 1 ≤ Y 2 ≤ m
- 1000 ≤ rectangular array within the element pixel the value of ≤ 1000
Sample input:
3 4 3
1 7 2 4
3 6 2 8
2 1 2 3
1 1 2 2
2 1 3 4
1 3 3 4
Sample output:
17
27
21
AC代码
#pragma GCC optimize(2) #include<bits/stdc++.h> using namespace std; inline int read() {int x=0,f=1;char c=getchar();while(c!='-'&&(c<'0'||c>'9'))c=getchar();if(c=='-')f=-1,c=getchar();while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();return f*x;} typedef long long ll; const int maxn = 1e3+10; int a[maxn][maxn]; int s[maxn][maxn]; int n,m,q; int main() { cin>>n>>m>>q; for(int i=1;i<=n;i++){ for(int j=1;j<=m;j++){ cin>>a[i][j]; s[i][j]=a[i][j]+s[i-1][j]+s[i][j-1]-s[i-1][j-1]; } } int x1,x2,y1,y2; int sum=0; for(int i=0;i<q;i++){ cin>>x1>>y1>>x2>>y2; sum=s[x2][y2]-s[x1-1][y2]-s[x2][y1-1]+s[x1-1][y1-1]; printf("%d\n",sum); } }