还是有点问题,但是思路应该是对的。
1.分析题目,是5种情况分别是被5除后余0,1,2,3,4,所以想到用switch语句,在多个固定选项中做出选择。
2.建立一个vector,用于存储输入的数字
3.再遍历整个vector,用for循环;当中每一个数字都是检测是否是5种情况中的一种。
4.最后输出的时候判断是否为空
case标签必须是整形常量表达式(case 0,这个0就是标签);
iomanip是setprecision的头文件,用来设置精度;
#include<iostream>
#include<string>
#include<vector>
#include<iomanip>
using namespace std;
int main()
{
vector<int> v;
int a, a1=0, a2=0, a3=0, a4=0, a5=0, count=0;
char c;
int flag=1;
while(cin >> a) //保存输入
{
v.push_back(a);
if((c=getchar())=='\n')
break;
}
for(int i=0;i<v.size();i++)
{
int rem = v[i]%5;
switch(rem)
{
case 0:
if(v[i]%2 == 0)
a1 += v[i];
break;
case 1:
a2 += v[i]*flag;
flag *= -1;
break;
case 2:
a3++;
break;
case 3:
a4 += v[i];
count++;
break;
case 4:
if(v[i]>a5)
a5 = v[i];
else
break;
}
}
if(a1 == 0)
cout << "N" << " ";
else cout << a1 << " ";
if(a2 == 0)
cout << "N" << " ";
else cout << a2 << " ";
if(a3 == 0)
cout << "N" << " ";
else cout << a3 << " ";
if(a4 == 0)
cout << "N" << " ";
else cout << setprecision(1) << a4/count << " ";
if(a5 == 0)
cout << "N";
else cout << a5;
return 0;
}