C :

#include<stdio.h>
#define max(a,b)	(a>b?a:b)
int m,n;
int a[110][110];
int main()
{
	scanf("%d %d",&m,&n);
	for(int i=1;i<=m;i++)
	for(int j=1;j<=n;j++)
	scanf("%d",&a[i][j]);
	for(int i=1;i<=m;i++)
	for(int j=1;j<=n;j++)
	a[i][j]=max(a[i-1][j],a[i][j-1])+a[i][j];
	printf("%d",a[m][n]);
	return 0;
}

C++ :

#include<iostream>
#include<cstring>
using namespace std;

int max(int a,int b){
	if(a>b)
		return a;
	else
		return b;
}

int main(){
	int n;//组数
	int b[101][101];//花生地中的花生数
	

		//初始化花生地 
		memset(b,0,sizeof(b)); 
		int x,y;//花生地的行列
		cin>>x>>y;//读入花生地的行列 
		
		//读入每块地种多少花生 
		//这里i和j从1开始，为的是下面判断i-1,j-1的时候边界值省得特殊处理 
		for(int i=1;i<=x;i++){
			for(int j=1;j<=y;j++){
				cin>>b[i][j];
			}
		} 
		
		//第推找最大值
		for(int i=1;i<=x;i++){
			for(int j=1;j<=y;j++){
				//从上面过来划算，还是从左边过来划算 
				b[i][j]+=max(b[i-1][j],b[i][j-1]);
			}
		} 
		
		cout<<b[x][y];
}

Java :

import java.util.Scanner;

public class Main {

	static int m,n;
	static int a[][] = new int[101][101];
	
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		m = sc.nextInt();
		n = sc.nextInt();
		int i,j;
		for(i=1;i<=m;i++) {
			for(j=1;j<=n;j++) {
				a[i][j] = sc.nextInt();
			}
		}
		for(i=1;i<=m;i++) {
			for(j=1;j<=n;j++) {
				if(i==1 && j==1) continue;
				else if(j>1 && i==1) a[i][j] = a[i][j-1] + a[i][j];
				else if(i>1 && j==1) a[i][j] = a[i-1][j] + a[i][j];
				else {
					a[i][j] = Math.max(a[i-1][j],a[i][j-1]) + a[i][j];
				}
			}
		}
		System.out.println(a[m][n]);
		sc.close();
	}
}

Python :

sr=input().split()
m=int(sr[0])
n=int(sr[1])
a=[[0 for i in range(n)] for j in range(m)]
s=[[0 for i in range(n)] for j in range(m)]
for i in range(m):
    sr=input().split()
    for j in range(n):
        a[i][j]=int(sr[j])
for i in range(m):
    for j in range(n):
        if(i>0 and j>0):
            s[i][j]=a[i][j]+max(s[i-1][j],s[i][j-1])
        elif(i>0):
            s[i][0]=a[i][0]+s[i-1][0]
        elif(j>0):
            s[0][j]=a[0][j]+s[0][j-1]
        else:
            s[0][0]=a[0][0]
print(s[m-1][n-1])