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 an example :
/* Note:Your choice is C IDE */#include "stdio.h"#define N 45int 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 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);}