C程序设计Week12晚上练习

本周仅仅进行一个程序，曾经的一个程序。

自己定义例如以下函数，输入n(n<46)个学生的姓名和成绩，顺序输出这n个学生的姓名和成绩，并输出最高成绩的姓名和成绩。预习struct结构体，思考怎样改进这一程序。

 //为count个学生输入姓名和成绩
 void getStudentsInfo(char names[][20], int scores[] , int count);
 void getStudentsInfo(char* names[], int scores[] , int count);
 //依次打印count个学生的姓名和成绩
void printStudentsInfo(char* names[], int scores[],int count);
//获取最高成绩的学生的index
int getIndexOfMaxScore(int scores[],int count);

知识点：

1 字符串数组用于处理多个人的姓名

2 数组作为函数的參数

the core code:

/* Note:Your choice is C IDE */
#include "stdio.h"

void inputStudents(char name[][20],int score[],int num);
void outputStudents(char name[][20],int score[],int num);
main()
{
    char name[45][20];
    int  score[45];

    inputStudents(name,score,2);
    outputStudents(name,score,2);

}

void inputStudents(char name[][20],int score[],int num){
	 int i ;
	 for(i=0;i<num;i++)
	   scanf("%s %d",name[i],&score[i]);
}
void outputStudents(char name[][20],int score[],int num){
	 int i ;
	 for(i=0;i<num;i++)
	  printf("%s %d",name[i],score[i]);
}

an example :

/* Note:Your choice is C IDE */
#include "stdio.h"
#define N 45

int inputSS(char names[][20],int score[],int num);
void printSS(char names[][20],int score[],int num);
void getMAX(char names[][20],int score[],int num);

main()
{
  char names[N][20];
  int score[N];
  int num=0;
  int choose;

  printf("What do you want to do: INPUT(1),OUTPUT(2),MAX(3),EXIT(0):");
  scanf("%d",&choose);
  do{
  	 switch(choose){
  	   case 1:
  	     num = inputSS(names,score,num);
  	     break;
  	   case 2:
  	     printSS(names,score,num);
  	     break;
  	   case 3:
  	     getMAX(names,score,num);
  	 }
  	 printf("\nWhat do you want to do: INPUT(1),OUTPUT(2),MAX(3),EXIT(0):");
  	 scanf("%d",&choose);
  	}while(choose != 0);

}
int inputSS(char names[][20],int score[],int num){
	int n,i;
	printf("\nThis Time, How many students do you want to input :");
	scanf("%d",&n);

	if((n+num)>N || n <1){
	  printf("not valid sum\n");
	  return -1;
	}
	printf("NOW INPUT AS ( NAME SCORE ):\n");
	for(i=0;i<n;i++){
	  printf("%d. ",i);
	  scanf("%s %d",&names[i+num],&score[i+num]);
	}
	printf("THIS TIME , INPUT IS OVER\n");
	return num+n;
}
void printSS(char names[][20],int score[],int num){
	int i ;
	if(num==0) {
		printf("NO STUDENTS NOW\n");
		return;
	}

	printf("\nNOW , THE STUDENTS SCORES AS FOLLOWS \n");

	for(i = 0 ;i<num;i++)
	  printf("%2d. name:%10s score:%3d\n",i,names[i],score[i]);
}
void getMAX(char names[][20],int score[],int num){

	int i ,max_index,max_score;

	if(num==0) {
		printf("NO STUDENTS NOW\n");
		return;
	}

	max_index=0;
	max_score=score[0];

	for(i = 1 ;i<num;i++)
	  if( score[i] > max_score ){
	  	 max_score = score[i];
	  	 max_index = i;
	  }

	printf("The Top Score is %d  by %s \n",score[max_index],names[max_index]);

}

=====================华丽的切割线====================================================

有非常多知识点须要大家复习。以下是一些比較cute的程序。弄懂啊弄懂

//0
#include "stdio.h"
void main(){
  int i=0, a[]={3,4,5,4,3};
  do
  {
      a[i]++;
   }while(a[++i]<5);

  for(i=0;i<5;i++)
     printf("%d",a[i]) ;
}

//1
#include "stdio.h"
void main(){
   int a = 7;
   int b = 8;
   printf ( "a&b = %d\n",a&b);
   printf( "a&&b = %d\n",a&&b);
}
//2
#include "stdio.h"
void main(){
   int i ;
   for(int i = 0; i<4; i++){
      if( i==2)
         break;
      printf("%d ",i);
   }
  printf("\n");
 for(int i = 0; i<4; i++){
      if( i==2)
         continue;
      printf("%d ",i);
   }
   printf("\n");
}
//3
#include "stdio.h"
void main(){
  int sum=0,item=0;
  while(item<7){
    item++;
    sum+=item;
    if(sum==7)
      break;  }
  printf("%d\n",sum);
}
//4
#include "stdio.h"
void main(){
  int a[]={1,2,3,4,5,6,7,8};
  int i,x ,*p;
  x=1;
  p=&a[3];
  for( i=0; i<3; i++ )
     x *= *(p+i);
  printf("x=%d\n",x);
}
//5
#include "stdio.h"
void main(){
  int i=5,x=1;
  for(;i<5;i++) x=x+1;
  printf("%d\n",x);
}
//6
#include "stdio.h"
void main(){
    int x,y;
    for (x=0, y=0 ; (y!=123) &&(x<4); x++)
       y++;
    printf("x=%d,y=%d\n",x,y);
}
//7
#include "stdio.h"
void main(){
 int a[7]={3,4,5,6,7,8,9};
 int *p,*q;
 int i,x;
 p=&a[0];
 q=&a[6];
 for (i=0;i<3;i++)
   if(*(p+i)==*(q-i) )
      x=*(p+i)*2;
}
//8
#include "stdio.h"
void main(){
  int a[5][5];
  printf("&a[3][2]-a=%d\n",&a[3][2]-a);
}
//9
#include "stdio.h"
void main(){
  int i=2,n=2;
  for(;i<5;i++){
     continue;
     n=n+i;
   }
  printf("%d\n",n);
}