61.
#include
#include //setprecision所需的库
using namespace std;
int main()
{
int n, age, sum = 0;
cin >> n;
for (int i = 1; i <= n; i++) // 循环n次
{
cin >> age; // 每循环一次输入一个学生的年龄
sum += age; // 求学生的年龄之和
}
double aver = 1.0 * sum / n; // 求平均年龄 1.0*后int型就变成浮点型了
cout << fixed << setprecision(2) << aver;
return 0;
}
结果:
62.
#include
#include
using namespace std;
int main()
{
int n, t, sum = 0;
cin >> n;
for (int i = 1; i <= n; i++)
{
cin >> t; // 每循环一次输入一个数据
sum += t; // 求和
}
double aver = 1.0 * sum / n; // 求均值
cout << fixed << setprecision(5);
cout << sum << ' ' << aver;
return 0;
}
结果:
63.
#include
using namespace std;
int main()
{
int n, score, Max = 0;
cin >> n;
for (int i = 1; i <= n; i++) //循环n次
{
cin >> score;
if (score > Max) Max = score; //如果score大于Max, 就将score的值赋予Max
}
cout << Max;
return 0;
}
结果:
64.
#include
using namespace std;
int main()
{
int n, Max = 0, Min = 1000, t; //Max初始值为0 Min初始值为1000
cin >> n;
for (int i = 1; i <= n; i++) //循环n次
{
cin >> t;
if (t > Max) Max = t; // 找最大值
if (t < Min) Min = t; // 找最小值
}
cout << Max - Min; //计算最大值和最小值之间的差
return 0;
}
结果:
65.
#include
using namespace std;
int main()
{
int n, Jin, Yin, Tong;
int JinSum = 0, YinSum = 0, TongSum = 0, sum;
cin >> n;
for (int i = 1; i <= n; i++) // 循环n次
{
cin >> Jin >> Yin >> Tong; // 输入一天获得的金银铜牌数
JinSum += Jin; // 金牌求和
YinSum += Yin; // 银牌求和
TongSum += Tong; // 铜牌求和
}
sum = JinSum + YinSum + TongSum; // 求得总牌数
cout << JinSum << ' ' << YinSum << ' ' << TongSum << ' ' << sum;
return 0;
}
结果:
66.
#include
using namespace std;
int main()
{
int m, n, sum = 0;
cin >> m >> n;
for (int i = m; i <= n; i++) //m和n之间遍历
if (i % 2 != 0) // 如果是奇数
sum += i; // 则求和
cout << sum;
return 0;
}
结果:
67.
#include
using namespace std;
int main()
{
int m, n, sum = 0;
cin >> m >> n;
for (int i = m; i <= n; i++)
if (i % 17 == 0) // 如果能被17整除
sum += i; // 则求和
cout << sum;
return 0;
}
结果:
68.
#include
using namespace std;
int main()
{
int k, t, one = 0, five = 0, ten = 0;
cin >> k;
for (int i = 1; i <= k; i++) //循环k次
{
cin >> t;
switch (t)
{
case 1:one++;break; //如果t=1 one增加1
case 5:five++;break; //如果t=5 five增加1
case 10:ten++; //如果t=10 ten增加1
}
}
cout << one << '
' << five << '
' << ten;
return 0;
}
结果:
69.
#include
using namespace std;
int main()
{
int N, m, t, cnt = 0;
cin >> N >> m;
for (int i = 1; i <= N; i++) //循环N次
{
cin >> t;
if (t == m) cnt++; //如果t==m cnt增加1
}
cout << cnt;
return 0;
}
结果:
70.
#include
using namespace std;
int main()
{
int a, n, power = 1;
cin >> a >> n;
for (int i = 1; i <= n; i++) //循环n次
power *= a; // 相当于power=power*a
cout << power;
return 0;
}
结果:
页面更新:2024-06-11
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