中考查分的网站:https://49.4.5.169:8018/jzstu

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中考查分的网站:https://49.4.5.169:8018/jzstu(共9篇)

中考查分的网站:https://49.4.5.169:8018/jzstu（一）：

#include
#include
typedef struct stu{
int id;
char name[40];
int mark[4];
float total;
} STU;
int id;
int m1,m2,m3,m4;
char str[40];
void main()
{
int i,j,N=0;
FILE *fin, *fout;
STU tmp;
STU *s;
fin = fopen("score.txt","r");
while (1){
if ( fscanf(fin,"%*d %*s %*d %*d %*d %*d") ==EOF) break;
N++;
}
fclose(fin);
s = (STU *) malloc(N * sizeof(STU));
if (!s) {
printf("no enough memory for STU\n"); exit(0);
}
fin = fopen("score.txt","r");
for (i=0;ifscanf(fin,"%d %s %d %d %d %d",&id,&str[0],&m1,&m2,&m3,&m4);
s[i].id = id;
strcpy(&s[i].name[0],str);
s[i].mark[0]= m1;
s[i].mark[1]= m2;
s[i].mark[2]= m3;
s[i].mark[3] = m4;
s[i].total = 0.2*m1 + 0.6*m2 + 0.5 * m3 + 0.15*m4;
}
fclose(fin);
for (i=0;ifor (j=i+1;jif (s[j].total > s[i].total){
tmp = s[j]; s[j]=s[i];s[i]=tmp;
};
};
};
fout = fopen("score_Q.txt","w");
for(i=0;ifprintf(fout,"%d %s %d %d %d %d\n",s[i].id,s[i].name,s[i].mark[0],s[i].mark[1],s[i].mark[2],s[i].mark[3]);
}
fclose(fout);
exit(0);
}

中考查分的网站:https://49.4.5.169:8018/jzstu（二）：

#include
#include
#include
using namespace std;
struct Student
{char name[20];
int num;
int a[4];
int rz;
float rj;
}stu[3]={{"Wang",101,55,57,80,96,0,0},{"Li",102,91,86,93,95,0,0},{"Zhao",103,82,61,49,52,0,0}};
struct a
{
int kz;
float kj;
}ke[4]={0,0,0,0,0,0,0,0};
void xs()
{cout <<<"单片机"<cout <<"--------------------------------------------------------------------------------";}
float avg(struct Student stu[],struct a ke[])
{
int i,j;
for(i=0;i<3;i++)
for(j=0;j<4;j++)
stu[i].rz=stu[i].rz+stu[i].a[j];
for(i=0;i<3;i++)
stu[i].rj=stu[i].rz/4.0;
for(i=0;i<4;i++)
for(j=0;j<3;j++)
ke[i].kz=ke[i].kz+stu[j].a[i];
for(i=0;i<4;i++)
ke[i].kj=ke[i].kz/3.0;
return 1;
}
float Avg(struct Student stu[],struct a ke[])
{
int i;
avg(stu,ke);
for(i=0;i<3;i++)
cout << cout <<"课程总分"< for(i=0;i<4;i++)
cout < cout <<"\b"< for(i=0;i<4;i++)
cout < cout <return 1;}
void highest(struct Student stu[])
{
int i,j,max=0,max_i,max_j;
for(i=0;i<3;i++)
for(j=0;j<4;j++)
if(stu[i].a[j]>max)
{max=stu[i].a[j];
max_i=i;max_j=j;}
cout <<"最高分为"< cout <<"其对应的学生是："< cout <<"其对应的课程是：";
if(max_j==0)cout <<"C++"< else if(max_j==1)cout <<"数据结构"< else if(max_j==2)cout <<"单片机"< else cout <<"英语"< cout <}
void bjg(struct Student stu[])
{
int i,j,s[3]={0,0,0};
string r[3];
for(i=0;i<3;i++)
for(j=0;j<4;j++)
if(stu[i].a[j]<60)s[i]++;
for(i=0;i<3;i++)
if(s[i]==2||s[i]>2)r[i]="否";
else r[i]="是";
cout <<<"单片机"<cout <<"--------------------------------------------------------------------------------";
for(i=0;i<3;i++)
cout <<}
void biger(struct Student stu[])
{
int i,j,s[3]={0,0,0},s1[3]={0,0,0};
string c[3],r[3];
for(i=0;i<3;i++)
for(j=0;j<4;j++)
if(stu[i].a[j]>85||stu[i].a[j]==85)s[i]++;
for(i=0;i<3;i++)
if(stu[i].rj==90||stu[i].rj>90||s[i]==4)c[i]="是";
else c[i]="否";
for(i=0;i<3;i++)
for(j=0;j<4;j++)
if(stu[i].a[j]<60)s1[i]++;
for(i=0;i<3;i++)
if(s1[i]==2||s1[i]>2)r[i]="否";
else r[i]="是";
cout <<<"单片机"<cout <<"--------------------------------------------------------------------------------";
for(i=0;i<3;i++)
cout <<}
void pjfc(struct Student stu[],struct a ke[])
{
int i;
float sum=0,sum1=0,s=0,s1=0;
avg(stu,ke);
for(i=0;i<3;i++)
sum=sum+stu[i].rj*stu[i].rj;
for(i=0;i<3;i++)
sum1=sum1+stu[i].rj;
cout <<"学生成绩平均方差是"< for(i=0;i<4;i++)
s=s+ke[i].kj*ke[i].kj;
for(i=0;i<4;i++)
s1=s1+ke[i].kj;
cout <<"课程平均方差是"< cout <}
void jm()
{cout <cout <<"1.输出学生基本成绩信息"<cout <<"2.输出最高分对应的学生和课程"<cout <<"3.如果两门以上（包括两门）课程不及格，输出及格为否，否则为是，输出学生信息"<cout <<"4.如果平均成绩在90分以上（包括90）或全部成绩在85分以上（包括85），输出优秀为是，否则为否，输出学生信息"<cout <<"5.求平均方差"<cout <<"6.退出"<}
int main()
{
int choice;
char s;
cout <<"**********************************"<cout <<" 欢迎使用 "<cout <<"**********************************"<jm();
while(1)
{Student stu[3]={{"Wang",101,55,57,80,96,0,0},{"Li",102,91,86,93,95,0,0},{"Zhao",103,82,61,49,52,0,0}};
a ke[4]={0,0,0,0,0,0,0,0};
cout <<"是否继续（y/n）";
cin >>s;
if(s=="y")
{jm();
cout <<"请选择（1-6）：";
cin >>choice;}
else return 0;
switch(choice)
{
case 1:

xs();
Avg(stu,ke);
break;
case 2:

highest(stu);
break;
case 3:

bjg(stu);

break;
case 4:
biger(stu);
break;
case 5:
pjfc(stu,ke);
break;
case 6:
return 0;
}}}