解题思路:
利用输入输出两个函数以及循环计数
注意事项:
注意计数
参考代码:
#include<iostream>
using namespace std;
struct student
{
string num;
string name;
int cj1,cj2,cj3;
};
struct student s[100];
void input(int n)
{
for (int i = 0; i < n; i++)
{
cin >> s[i].num >> s[i].name >> s[i].cj1>>s[i].cj2>>s[i].cj3;
}
}
void output(int n)
{
int average[3] = { 0,0,0 };
int sum[100];
int shu = 0;
for (int i = 0; i < n; i++)
{
average[0] += s[i].cj1;
average[1] += s[i].cj2;
average[2] += s[i].cj3;
sum[i] = s[i].cj1 + s[i].cj2 + s[i].cj3;
if (i >= 2)
{
if (sum[i > sum[i - 1]])
{
shu = i;
}
}
}
cout << average[0] / n << " " << average[1] / n << " " << average[2] / n << endl;
cout << s[shu].num << " " << s[shu].name << " " << s[shu].cj1 << " " << s[shu].cj2 << " " << s[shu].cj3 << endl;
}
int main()
{
int n;
cin >> n;
input(n);
output(n);
return 0;
}
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