Soluciones y Explicación de los problemas ACM ... - ICPC Bolivia

Explicaciones a los problemas South American Regionals 2008 25
Programa Java que soluciona el problema
import java.util.Scanner;
import static java.lang.Math.max;
class candy {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int M, N;
while(true){
M = in.nextInt();
N = in.nextInt();
if(M == 0 && N == 0) break;
int[][] D = new int[M][N + 2];
for (int i = 0; i < M; i++) {
for (int j = 0; j < N; j++) {
D[i][j + 2] = in.nextInt();
}
}
for (int i = 0; i < M; i++) {
for (int j = 2; j < (N + 2); j++) {
D[i][j] = max(D[i][j - 1], D[i][j] + D[i][j - 2]);
}
}
int[] sol = new int[M + 2];
for (int i = 0; i < M; i++) {
sol[i + 2] = D[i][N + 1];
}
for (int i = 2; i < (M + 2); i++) {
sol[i] = max(sol[i - 1], sol[i] + sol[i - 2]);
}
System.out.println(sol[M + 1]);
}
}
}
Programa C que soluciona el problema
#include
#define max(_x,_y) ((_x) > (_y) ? (_x) : (_y))
int main(void)
{
int N,M,i,j,a,b,c,A,B,C;