PAT Basic level practice 12
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1012. 数字分类 (20)
时间限制
50 ms
内存限制
65536 kB
代码长度限制
8000 B
判题程序
Standard
作者
CHEN, Yue
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:
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
C++:
#include<iostream>
using namespace std;#include<cstdlib>
#include<vector>
#include<iomanip>
int main()
{
int time;
cin >> time;
int temp;
int sum1 = 0, sum2 = 0, cnt3 = 0, cnt2 = 0, sum4 = 0,cnt4 = 0;
float aver4;
vector<int> vec1;
int *a = new int[time];
for (int i = 0; i < time; i++)
{
cin >> a[i];
}
for (int i = 0; i < time; i++)
{
temp = a[i] % 5;
switch (temp)
{
case 0://A1
{
if (a[i] % 2 == 0)
sum1 += a[i];
break;
}
case 1://A2
{
if (cnt2 % 2 == 0)
{
sum2 += a[i];
cnt2++;
}
else
{
sum2 -= a[i];
cnt2++;
}
break;
}
case 2://A3
{
cnt3++;
break;
}
case 3://A4
{
sum4 += a[i];
cnt4++;
break;
}
case 4:
{
vec1.push_back(a[i]);
break;
}
default:
break;
}
}
aver4 = float(sum4) / float(cnt4);
int max5 = 0;
int size = vec1.size();
for (int i = 0; i < size; i++)
{
if (max5 < vec1[i])
max5 = vec1[i];
}
if (sum1 == 0)
cout << "N ";
else
cout << sum1 << " ";
if (cnt2 == 0)
cout << "N ";
else
cout << sum2 << " ";
if (cnt3 == 0)
cout << "N ";
else
cout << cnt3 << " ";
if (cnt4 == 0)
cout << "N ";
else
cout << setiosflags(ios::fixed) << setprecision(1) << aver4<<" ";
if (max5 == 0)
cout << "N";
else
cout << max5;
system("Pause");
}
评测结果
测试点
0 0
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