PAT 乙级 1012. 数字分类 (20) C++版

给定一系列正整数,请按要求对数字进行分类,并输出以下5个数字:  

  • A1 = 能被5整除的数字中所有偶数的和;
  • A2 = 将被5除后余1的数字按给出顺序进行交错求和,即计算n1-n2+n3-n4…;
  • A3 = 被5除后余2的数字的个数;
  • A4 = 被5除后余3的数字的平均数,精确到小数点后1位;
  • A5 = 被5除后余4的数字中最大数字。

输入格式:

每个输入包含1个测试用例。每个测试用例先给出一个不超过1000的正整数N,随后给出N个不超过1000的待分类的正整数。数字间以空格分隔。

输出格式:

对给定的N个正整数,按题目要求计算A1~A5并在一行中顺序输出。数字间以空格分隔,但行末不得有多余空格。 若其中某一类数字不存在,则在相应位置输出“N”。

输入样例1:

13 1 2 3 4 5 6 7 8 9 10 20 16 18

输出样例1:

30 11 2 9.7 9

输入样例2:

8 1 2 4 5 6 7 9 16

输出样例2:

N 11 2 N 9

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#include <iostream>
#include <cstdio>

using namespace std;

void getA1(int *number, int n) {
int sum = 0;
bool isEnter = false;
for (int i = 0; i < n; i++) {
if (number[i] % 5 == 0 && number[i] % 2 == 0) {
sum += number[i];
isEnter = true;
}
}

if (!isEnter) {
cout << "N ";
} else {
cout << sum << " ";
}
}

void getA2(int *number, int n) {
int sign = 1;
int sum = 0;
bool isEnter = false;
for (int i = 0; i < n; i++) {
if (number[i] % 5 == 1) {
sum += sign * number[i];
sign = -sign;
isEnter = true;
}
}

if (isEnter == false) {
cout << "N ";
} else {
cout << sum << " ";
}
}

void getA3(int *number, int n) {
int cnt = 0;
for (int i = 0; i < n; i++) {
if (number[i] % 5 == 2) {
cnt++;
}
}

if (cnt == 0) {
cout << "N ";
} else {
cout << cnt << " ";
}
}

void getA4(int *number, int n) {
int cnt = 0, sum = 0;
for (int i = 0; i < n; i++) {
if (number[i] % 5 == 3) {
sum += number[i];
cnt++;
}
}

if (cnt == 0) {
cout << "N ";
} else {
printf("%.1f ", sum * 1.0 / cnt);
}
}

void getA5(int *number, int n) {
int max = 0;
for (int i = 0; i < n; i++) {
if (number[i] % 5 == 4) {
if (max < number[i]) {
max = number[i];
}
}
}

if (max == 0) {
cout << "N";
} else {
cout << max;
}
}

int main() {
int n;
cin >> n;

int *number = new int[10001];
for (int i = 0; i < n; i++) {
cin >> number[i];
}

getA1(number, n);
getA2(number, n);
getA3(number, n);
getA4(number, n);
getA5(number, n);

delete[] number;
}