91. The establishment of a dynamic array, student performance input 5, wherein further with a function to check whether or below 60, and outputs the results failed
#include <stdio.h>
#include <stdlib.h>
void check(int *p) //打印成绩<60的
{
int i;
for(i=0; i<5; i++)
if(*(p+i)<60)
printf("%d ",*(p+i));
printf("\n");
}
int main()
{
int *p1,i;
p1=(int *)malloc(5*sizeof(int)); //动态开辟空间
for(i=0; i<5; i++)
scanf("%d",p1+i); //p1+i为地址
check(p1);
return 0;
}
operation result:
92. Verify: Goldbach conjecture: an even number of not less than 6 can be expressed as the sum of two primes. Example: 34 = 3 + 31
#include <stdio.h>
#include <math.h>
int prime(int m) //判断是否为素数
{
int i,k;
k=sqrt(m);
for(i=2; i<=k; i++)
if(m%i==0)
break;
if(i>k)
return 1;
else
return 0;
}
void godbaha(int n)
{
int a,b;
for(a=3; a<=n/2; a+=2) //a为第一个素数，一定不大于n/2
{
if(prime(a))
{
b=n-a; //判断第二个是否为素数
if(prime(b))
printf("%d=%d+%d\n",n,a,b);
}
}
}
int main()
{
int number;
printf("输入一个不小于6的偶数：");
scanf("%d",&number);
godbaha(number);
printf("\n");
return 0;
}
operation result:
93. Enter the 10 student's name, student number and scores, failing which the person's name, student number and output results
#include <stdio.h>
int main()
{
char name[10][10]; //姓名
int no[10]; //学号
float score[10]; //成绩
int i;
for(i=0; i<10; i++) //三者通过数组下标相联系
scanf("%s %d %f",name[i],&no[i],&score[i]);
printf("打印成绩不及格的学生：\n");
for(i=0; i<10; i++)
if(score[i]<60)
printf("%s %d %.2f\n",name[i],no[i],score[i]);
return 0;
}
operation result:
94. The definition of the structure, membership information Print
#include <stdio.h>
struct Student
{
long int num; //学号
char name[20]; //姓名
char sex; //性别
char addr[30]; //地址
} s1= {10101,"Li Lin",'M',"123 Beijing Road"};
int main()
{
printf("No:%ld\nname:%s\nsex:%c\naddress:%s\n",s1.num,s1.name,s1.sex,s1.addr); //输出
return 0;
}
operation result:
95. There are three candidates, each voter can only vote one person, compile a program, get voting results
#include <stdio.h>
#include <string.h>
struct Person
{
char name[30]; //姓名
int count; //票数
} leader[3]= {"Li",0,"Zhang",0,"Sun",0};
int main()
{
int i,j;
char leader_name[30];
for(i=0; i<10; i++) //依次输入10个票选
{
scanf("%s",leader_name);
for(j=0; j<3; j++) //输入一个判断一次
if(strcmp(leader_name,leader[j].name)==0) //将候选人票数加1
leader[j].count++;
}
printf("Result:\n"); //输出各候选人票数
for(i=0; i<3; i++)
printf("%5s:%d\n",leader[i].name,leader[i].count);
printf("\n");
return 0;
}
operation result:
96. There are n number of students, according to results of the output level of student information
#include <stdio.h>
struct Student //学生结构体
{
int num;
char name[20];
float score;
};
int main()
{
struct Student stu[5]= {{10101,"Zhang",78},{10102,"Wang",98.5},{10106,"Li",86},{10108,"Ling",73.5},{10110,"Sun",100}};
struct Student temp;
const int n=5;
int i,j;
for(i=0; i<n-1; i++) //根据成绩进行冒泡排序
for(j=0; j<n-i-1; j++)
if(stu[j].score>stu[j+1].score)
{
temp=stu[j]; //结构体可以直接进行赋值
stu[j]=stu[j+1];
stu[j+1]=temp;
}
for(i=0; i<n; i++) //输出排序后的结果
printf("%6d %8s %6.2f\n",stu[i].num,stu[i].name,stu[i].score);
printf("\n");
return 0;
}
operation result:
97. There are n structure variable containing student number, name and three scores. Requirements: Output highest average score of student information
#include <stdio.h>
#define N 3
struct Student //学生结构体
{
int num; //学号
char name[20]; //姓名
float score[3]; //三门成绩
float aver; //平均成绩
};
struct Student max(struct Student stu[]) //求平均成绩最高的学生
{
int i,m=0;
for(i=0; i<N; i++)
if(stu[i].aver>stu[m].aver)
m=i;
return stu[m];
}
void input(struct Student stu[]) //输入学生信息及求平均成绩
{
int i;
for(i=0; i<N; i++)
{
scanf("%d %s %f %f %f",&stu[i].num,stu[i].name,&stu[i].score[0],&stu[i].score[1],&stu[i].score[2]);
stu[i].aver=(stu[i].score[0]+stu[i].score[1]+stu[i].score[2])/3.0;
}
}
void print(struct Student stud) //打印学生信息
{
printf("%d\n%s\n%5.1f\n%5.1f\n%5.1f\n%6.2f\n",stud.num,stud.name,stud.score[0],stud.score[1],stud.score[2],stud.aver);
}
int main()
{
struct Student stu[N],*p=stu;
input(p); //直接用stu相同
printf("平均成绩最高的学生：\n");
print(max(p)); //直接用stu相同
return 0;
}
operation result:
98. A variable pointer to a structure
#include <stdio.h>
#include <string.h>
struct Student
{
long num;
char name[20];
};
int main()
{
struct Student stu_1;
struct Student *p; //定义指向结构体变量的指针
p=&stu_1;
stu_1.num=10101;
strcpy(stu_1.name,"Li Ming"); //下述三种方法结果相同
printf("直接调用：\n");
printf("No:%ld\nname:%s\n",stu_1.num,stu_1.name);
printf("结构体变量指针调用用法：\n");
printf("No:%ld\nname:%s\n",(*p).num,(*p).name); //用*p调用
printf("No:%ld\nname:%s\n",p->num,p->name); //用p->调用
return 0;
}
operation result:
99. The structure of the variable pointers Usage
#include <stdio.h>
struct Student
{
long num;
char name[20];
};
int main()
{
struct Student stu[3]= {{10101,"Li Min"},{10102,"Zhang Zi"},{10103,"Wang Min"}};
struct Student *p;
for(p=stu; p<stu+3; p++) //和普通数组相同用法
printf("%5ld %-20s\n",p->num,p->name);
return 0;
}
operation result:
100. write a function to establish a three-way dynamic list of student data
#include <stdio.h>
#include <stdlib.h> //malloc
struct Student
{
long num;
float score;
struct Student *next;
};//定义学生结构体
struct Student * Create(void) //创建学生链表
{
struct Student *head=NULL,*p1,*p2;
p1=p2=(struct Student *)malloc(sizeof(struct Student));
while(scanf("%ld,%f",&p1->num,&p1->score)&&p1->num!=0&&p1->score!=0) //输入学生信息直到输入"0,0"
{
if(head==NULL) //一开始链表为空
head=p1;
else //链表不为空
p2->next=p1;
p2=p1; //p2为循环指针
p1=(struct Student *)malloc(sizeof(struct Student));
}
p2->next=NULL;
return head;
}
void print(struct Student *head) //打印学生链表
{
struct Student *p;
p=head;
if(p!=NULL)
while(p!=NULL)
{
printf("%ld %5.1f\n",p->num,p->score);
p=p->next;
}
}
int main()
{
struct Student *pt;
pt=Create();
printf("输出学生数据：\n");
print(pt);
return 0;
}
operation result:
