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public class Customer { | |
private Long ecleticaId; | |
private String name; | |
private String phone; | |
private String cep; | |
@Setter |
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A produção foi igual a 2.769 milhões de toneladas em terra não irrigada tendo um custo de $9 x 1 milhão de tonelada, e 3 milhões em terra irrigada tendo um custo de $11 x 1 milhão de tonelada. Logo o custo será de ((9 * 2.769) + (11 * 3)) = 24.92 + 33 = $ 57.92 milhões. |
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#include<stdio.h> | |
void Quick(int vetor[3], int inicio, int fim) { | |
int pivo, aux, i, j, meio; | |
i = inicio; | |
j = fim; | |
meio = (int) ((i + j) / 2); |
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BinaryTree* InsertNode(BinaryTree *bt, int value) { | |
if(bt == NULL) { | |
BinaryTree *bt = (BinaryTree*) malloc(sizeof(BinaryTree)); | |
bt->value = value; | |
bt->left = NULL; | |
bt->right = NULL; | |
return (bt); | |
} else { | |
BinaryTree *aux = (BinaryTree*) malloc(sizeof(BinaryTree)); | |
aux = bt; |
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typedef struct pilha { | |
int size; | |
int itens[10]; | |
} Pilha; | |
Pilha* criapilha() { | |
Pilha *pilha = (Pilha*) malloc(sizeof(Pilha)); |