題目連結

思路簡介

     每類數字設計一個函式，根據函式返回值來判斷該類數字有沒有，以及存在時數字是多少。

程式碼

#include <stdlib.h>
#include <stdio.h>


int A1(int a[], int num){
	int i = 0, temp = 0;
	while (i < num){
		if (a[i] % 5 == 0 && a[i] % 2 == 0)
			temp += a[i];
		i++;
	}
	return temp;
}

bool A2(int a[], int num, int * temp){
	int i, flag;
	flag = i = *temp = 0;
	while (i < num){
		if (a[i] % 5 == 1){
			flag++;
			if (flag % 2 == 1)
				*temp += a[i];
			else
				*temp -= a[i];
		}
		i++;
	}
	if (flag == 0)
		return false;
	else
		return true;
}
int A3(int a[], int num){
	int i = 0, flag = 0;
	while (i < num){
		if (a[i] % 5 == 2)
			flag++;
		i++;
	}
	return flag;
}
double A4(int a[], int num){
	int i = 0, flag = 0;
	double temp = 0;
	while (i < num){
		if (a[i] % 5 == 3){
			temp += a[i];
			flag++;
		}
		i++;
	}
	if (flag != 0)
		temp /= flag;
	return temp;
}

int A5(int a[], int num){
	int i = 0, max = 0;
	while (i < num){
		if (a[i] % 5 == 4)
			if (a[i] > max)
				max = a[i];
		i++;
	}
	return max;
}

void print(int array[],int _N){
	int num1;
	double num2;

	num1 = A1(array, _N);
	if (num1 == 0)
		printf("N ");
	else
		printf("%d ", num1);

	if (A2(array, _N, &num1))
		printf("%d ", num1);
	else
		printf("N ");

	num1 = A3(array, _N);
	if (num1 == 0)
		printf("N ");
	else
		printf("%d ", num1);

	num2 = A4(array, _N);
	if (num2 == 0)
		printf("N ");
	else
		printf("%.1f ", num2);

	num1 = A5(array, _N);
	if (num1 == 0)
		printf("N");
	else
		printf("%d", num1);
}

int main(){
	int array[1000] = {0},
	    N,j;

	scanf("%d", &N);
	j = 0;
	while (j <N - 1){
		scanf("%d ", &array[j]);
		j++;
	}
	scanf("%d", &array[j]);

	print(array,N);
	return 0;
}