分享一个c++题解

#include<bits/stdc++.h>
using namespace std;
#define int long long
int a[1010][1010],b[1010][1010];//a数组存储原来数字 b数组用来存每行的前缀和
signed main(){
 int n,m,q;
 cin>>n>>m>>q;
 for(int i=1;i<=n;i++){
    for(int j=1;j<=m;j++){
        cin>>a[i][j];
     if(j!=1)   b[i][j]+=b[i][j-1]+a[i][j];//前缀和
     else {
        b[i][j]=a[i][j];//第一列就等于原本的
     }
    }
 }
 for(int k=1;k<=q;k++){
    int x1,y1,x2,y2,ans=0;
    cin>>x1>>y1>>x2>>y2;
    //cout<<b[x1][y2]<<" "<<b[x1][y1]<<" " <<b[x2][y2]<<" "<<b[x2][y1]<<endl;
    for(int i=x1;i<=x2;i++){
        ans+=b[i][y2]-b[i][y1-1];
    }
    cout<<ans<<endl;
 }

}

JAVA AC代码:

import java.util.*;
import java.util.StringTokenizer;
import java.io.*;

public class Main {

	public static void main(String[] args) {
		int n=in.nextInt();
		int m=in.nextInt();
		int q=in.nextInt();
		int[][] arr=new int[n+10][m+10];
		int[][] sum=new int[n+10][m+10];
		for(int i=1;i<=n;i++){
			for(int j=1;j<=m;j++){
				arr[i][j]=in.nextInt();
				sum[i][j]=arr[i][j]+sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1];
			}
		}
		while(q-->0){
			int x1=in.nextInt(),y1=in.nextInt(),x2=in.nextInt(),y2=in.nextInt();
			int ans=sum[x2][y2]-sum[x1-1][y2]-sum[x2][y1-1]+sum[x1-1][y1-1];
			out.println(ans);
		}
		out.close();
	}
	
	

	static FastReader in=new FastReader();
	static PrintWriter out=new PrintWriter(System.out);
	static class FastReader{
		static BufferedReader br;
		static StringTokenizer st;
		FastReader(){
			br=new BufferedReader(new InputStreamReader(System.in));
		}
		String next() {
			String str="";
			while(st==null||!st.hasMoreElements()) {
				try {
					str=br.readLine();
				}catch(IOException e) {
					throw new RuntimeException(e);
				}
				st=new StringTokenizer(str);
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
	
}