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。
输入描述:
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。
输出描述:
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”。
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
int N;
cin >> N;
int a[1000];
memset(a, 0, sizeof(a));
int i = 0;
int A1 = 0;
int A2 = 0;
int A3 = 0;
double A4 = 0;
int A5 = 0;
int flag = 1;
int count_A2 = 0;//计数,判断这一类数字是否存在
int count_A3 = 0;
int count_A4 = 0;
while (N--)
{
cin >> a[i];
switch (a[i] % 5)
{
case 0:
if (a[i] % 2 == 0)
{
A1 = A1 + a[i];
} break;
case 1:
A2 = A2 + flag * a[i];
flag = flag * (-1);
count_A2++; break;
case 2:
A3++;
count_A3++; break;
case 3:
A4 = A4 + a[i];
count_A4++; break;
case 4:
if (A5<a[i])
{
A5 = a[i];
} break;
}
i++;
}
(A1>0) ? (cout << A1<< " ") :(cout << "N ");
(count_A2>0) ? (cout << A2 << " ") :(cout << "N ");
(count_A3>0) ? (cout << count_A3 << " ") : (cout << "N ");
(count_A4>0) ? (cout << fixed << setprecision(1) << A4 / count_A4 << " ") : (cout << "N ");//定点式输出小数点后一位
(A5>0) ? (cout << A5 ) : (cout << "N");//末尾无空格
return 0;
}