数据结构课程设计 问题是:给出n个学生的m门成绩，每个学生的信息由学号、姓名以及各科成绩组成

class CStudent { int nNumber[5]; char name[5][20]; int nChinese[5]; int nMath[5]; int nEnglish[5]; int nAveChinese; int nAveMath; int nAveEnglish; int nSumChinese; int nSumMath; int nSumEnglish; public: CStudent(char name1[][20]) { for (int i = 0; i < 5; i++) { nNumber[i] = i + 1; strcpy(name[i], name1[i]); nChinese[i] = rand()%90 + 60; nMath[i] = rand()%90 + 60; nEnglish[i] = rand()%90 + 60; } nAveChinese = 0; nAveMath = 0; nAveEnglish = 0; nSumChinese = 0; nSumMath = 0; nSumEnglish = 0; } void PrintAll(); void PrintAveChinese(); void PrintAveMath(); void PrintAveEnglish(); friend void Print(CStudent student); }; void CStudent::PrintAll() { cout<<"学号\t"<<"姓名\t"<<"语文\t"<<"数学\t"<<"英语"<<endl; for (int i = 0; i < 5; i++) { cout<<nNumber[i]<<"\t"<<name[i]<<"\t"<<nChinese[i]<<"\t"<<nMath[i]<<"\t"<<nEnglish[i]<<endl; } } void Print(CStudent student) { for (int i = 0; i < 5; i++) { cout<<studentnNumber[i]<<"\t"<<studentname[i]<<"\t"; if (studentnChinese[i] >= 140) cout<<"优\t"; else if (studentnChinese[i] < 140 && studentnChinese[i] >= 120) cout<<"良\t"; else if (studentnChinese[i] < 120 && studentnChinese[i] >= 100) cout<<"中\t"; else if (studentnChinese[i] < 100 && studentnChinese[i] >= 90) cout<<"及格\t"; else cout<<"不及格\t"; if (studentnMath[i] >= 140) cout<<"优\t"; else if (studentnMath[i] < 140 && studentnMath[i] >= 120) cout<<"良\t"; else if (studentnMath[i] < 120 && studentnMath[i] >= 100) cout<<"中\t"; else if (studentnMath[i] < 100 && studentnMath[i] >= 90) cout<<"及格\t"; else cout<<"不及格\t"; if (studentnEnglish[i] >= 140) cout<<"优\t"; else if (studentnEnglish[i] < 140 && studentnEnglish[i] >= 120) cout<<"良\t"; else if (studentnEnglish[i] < 120 && studentnEnglish[i] >= 100) cout<<"中\t"; else if (studentnEnglish[i] < 100 && studentnEnglish[i] >= 90) cout<<"及格\t"; else cout<<"不及格\t"; cout<<endl; } } void CStudent::PrintAveChinese() { for (int i = 0; i < 5; i++) nSumChinese += nChinese[i]; nAveChinese = nSumChinese/5; cout<<"语文平均成绩："<<nAveChinese<<endl; } void CStudent::PrintAveMath() { for (int i = 0; i < 5; i++) nSumMath += nMath[i]; nAveMath = nSumMath/5; cout<<"数学平均成绩："<<nAveMath<<endl; } void CStudent::PrintAveEnglish() { for (int i = 0; i < 5; i++) nSumEnglish += nEnglish[i]; nAveEnglish = nSumEnglish/5; cout<<"英语平均成绩："<<nAveEnglish<<endl; } int _tmain(int argc, _TCHAR argv[]) { srand(time(NULL)); char name1[5][20] = {"SZW", "SDL", "LHW", "LH", "LJ"}; CStudent student(name1); //student = new CStudent; studentPrintAll(); studentPrintAveChinese(); studentPrintAveMath(); studentPrintAveEnglish(); Print(student); }

记得采纳啊

1d1-->d2-->d3-->d4(线形关系)

2语句的频度就是该语句执行的次数：

(1)执行了n-1次,即用线性阶表示就是o(n)

(2)执行了n-1次, 即用线性阶表示就是o(n)

(3)执行了n-1次,即用线性阶表示就是o(n)

(4)执行了(n+1)n/2次, 用平方阶表示就是o (nn)

3我是用的快速排序，也有其它的方法

#include <stdioh>

void getMax(int a,int b,int c){

int m;

if (a<b){

m=b;b=a;a=m;

}

if (a<c){

m=c;c=a;a=m;

}

if (b<c){

m=c;c=b;b=m;

}

printf("%d%d%d",a,b,c);

}

void main(){

int a;

int b;

int c;

printf("please input a,b,c");

scanf("%d%d%d",&a,&b,&c);

getMax(a,b,c);

}

4#include <stdioh>

int importw(){

int n;

int i;

int maxint=1000000000;

long max=1;

printf("请输入正整数n:\n");

scanf("%d",&n);

if (n<=0){

n=importw();

}

else if (n>20){

printf("输入的n大于存储长度，请您输入n小于20");

n=importw();

}

else{

if (n==1){

max=2;

}

else{

for (i=1;i<=n;i++){

max=maxi;

}

max=max2;

if (max>maxint||max<0){

printf("您输入的n值过大，请重新输入\n");

n=importw();

}

}

}

return n;

}//得到正确的n值

void accountw(){

int n;

n=importw();

int a[20];

int i;

int j;

if (n==1){

a[0]=2;

printf("值列表为:\n");

printf("%d",a[0]);

}

else {

a[0]=2;

for (i=1;i<n;i++){

int amass;

amass=amassi;

a[i]=amass2;

}

printf("结果列表:\n");

for (j=1;j<n;j++){

printf("%d\n",a[j]);

}

}

}//计算阶乘的值

void main(){

accountw();

}

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