编程,某歌手大赛有10位选手10位评委要求评委给10位选手打分去掉一个最低分去掉一个最高分求出平均分

2024-11-21 03:17:42
推荐回答(4个)
回答(1):

1、首先,定义一个整型变量num,保存评委的人数。

2、接着,定义5个实型变量,保存所打的分数、最低分、最高分、总分和平均分。

3、初始化最低分和最高分,最低分为1000分,最高分为0分。

4、接着,设置总分分数,值为0。

5、输入评委的人数,保存在变量num中。

6、然后,提示输入评委的打分。

7、执行for循环语句,条件为i小于等于num。

8、运行程序,输入各评委的打分后,就会计算出节目的最后得分。

回答(2):

#include 
#include
#include
using namespace std;
bool MinScore(double v1, double v2)
{
    return (v1 < v2);
}
int main(int argc, char *argv[])
{
    vector dvec;
    double score;
    cout << "Please input 10 scores: " << endl;
    while (cin >> score && dvec.size() != 10)
    {
        dvec.push_back(score);
    }
    stable_sort(dvec.begin(), dvec.end(), MinScore);
    for (int i = 0; i < 10; ++i)
    {
        cout << dvec[i] << " ";
    }
    cout << endl;
    vector::iterator it = dvec.begin()+1;
    while (it != dvec.end()-1)
    {
        score += *it++;
    }
    score /= (dvec.size()-2);
    cout << "Score = " << score << endl;
    return 0;
}


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回答(3):

include
include
double average(int i,flot max,flot min);
void main ()
flote a[10][10],double b[10];
int i,j;
flote max,min;
for(i=0;i<10;i++)
for(j=0;j<10;j++)
scanf("f%",a);//评分输入
for(i=0;i<10;i++)
for(j=0;j<10;j++)

max=(a[i][j]>a[i][j+1]);
min=(a[i][j]}
for(i=0;i<=10;i++)
{
b[i]=average(i);
printf("%.2f/n",b[i]);
}
double average(int i,flote max,flote min)
{
double m=0;
int j;
for(j=0;j<=10;j++)
m+=a[i][j];
m=(m-max-min)/8
return m;
}

回答(4):

我这个是C++的代码 里面含有标准库。
#include
#include
#include
#include
#include
#include
using namespace std;

class Person
{
public:
Person(){};
Person(string name, double score)
{
this->m_Name = name;
this->m_Score = score;
}

string m_Name;
double m_Score; //平均分

};

void creatPerson(vector&v,int n)
{

int score = 0;
for (int i = 0; i < n; ++i)
{
string name1 = to_string(i+1); //int类型转化为string类型
string name2 = "号选手";
name1 += name2;
Person p(name1, score);
v.push_back(p);
}

}

void printVector(vector&v)
{
for (vector::iterator it = v.begin(); it != v.end(); ++it)
{
cout << it->m_Name << "最终平均分数为:" << it->m_Score << endl;

}

}

void printVector2(vector&v)
{
for (vector::iterator it = v.begin(); it != v.end(); ++it)
{
cout << it->m_Name << " ";
}
cout << endl;
}

void setVector(vector&v,int m)
{
for (vector::iterator it = v.begin(); it != v.end(); ++it)
{
dequed;
cout << "已为" << it->m_Name <<"打了"<for (int i = 0; i < m; ++i)
{
int score = rand() % 41 + 60;
d.push_back(score);
}

for (deque::iterator dit = d.begin(); dit != d.end(); ++dit)
{
cout << *dit << " ";
}
cout << endl;

sort(d.begin(), d.end());

d.pop_front();
d.pop_back();

/*for (deque::iterator dit = d.begin(); dit != d.end(); ++dit)
{
cout << *dit << " ";
}
cout << endl;*/

int sum = 0;
for (deque::iterator dit = d.begin(); dit != d.end(); ++dit)
{
sum += *dit;
}
//cout << "sum=" << sum << "数量" << d.size() << endl;

double avg = sum*1.0 / d.size();

it->m_Score = avg;

}
cout << endl;
}

void sortScore(vector&v)
{
vector::iterator it = v.begin();

Person t;
for (unsigned int i = 0; i < v.size(); ++i)
{
for (unsigned int j = 0; j < v.size() - i - 1; ++j)
{
if (it[j].m_Score>it[j + 1].m_Score) //比较分数
{
t = it[j]; //更换整体
it[j] = it[j + 1];
it[j + 1] = t;
}
}
}

}

int main(void)
{
cout << "软件说明:" << endl;
cout << "该软件为N个评委为M个选手打分,这样每一个选手就收到N次打分,去掉一个最低分和一个最高分,算出选手的平均分,并为这些选手按平均分从低到高排序。" << endl;
cout << endl;
cout << "请输入选手的个数N=";
int n;
while (true)
{
char c = cin.peek();
if (c > '0'&& c <= '9')
{
cin >> n;
cout << "你设置的选手人数为:" << n << endl;
break;
}
cin.clear(); //重置标志位
cin.sync(); //清空缓冲区
cout << "你输入不正确,请重新输入" << endl;
}
getchar();
cout << endl;
cout << "请输入评委的个数M=";
int m;
while (true)
{
char c = cin.peek();
if (c > '0'&& c <= '9')
{
cin >> m;
cout << "你设置的评委人数为:" << m << endl;
break;
}
cin.clear(); //重置标志位
cin.sync(); //清空缓冲区
cout << "你输入不正确,请重新输入" << endl;
}
cout << endl;
cout << "请这"<cout << "(为了避免手动输入分数浪费时间,下面让计算机随机为选手打"<cout << endl;
srand((unsigned int)time(NULL)); //随机数种子
vectorv;
creatPerson(v,n);

//printVector(v);
setVector(v,m);

cout << endl;
cout << "去掉一个最低分,去掉一个最高分:"<printVector(v);

cout << endl;
cout << "选手成绩由低到高排序如下:" << endl;
sortScore(v);
printVector2(v);

system("pause");
return 0;
}