C语言基础代码

1、输入成绩，判断是否及格#include<stdio.h>voidmain(){floata,b,max;printf("pleaseentera:");scanf("%f",&a);if(a>=60)printf("及格\n");elseprintf("不及格\n");}2、输入两个学生a和b的成绩，输出其中高的成绩#include<stdio.h>voidmain(){floata,b,max;printf("pleaseenteraandb:");scanf("%f,%f",&a,&b); //数据输入时若无逗号一般用空格区分两输入值，本例一定要在两数之间加逗号if(a>=b)max=a;if(b>a)max=b;printf("max=%f\n",max);}3、要求按照百分制考试分数输出等级。#include"stdio.h"main(){intgrade;printf("inputgrade=")；scanf("％d"，&grade)；switch((int)(grade/10)){case10:case9:printf("A\n")； break；case8:printf("B\n")；break；case7:printf("C\n")；break；case6:printf("D\n")； break；default:printf("E\n")；}黑龙江大学计算机学院、软件学院 《程序设计基础》实验讲义黑龙江大学计算机学院、软件学院 《程序设计基础》实验讲义、求1+2+ +100的和，分别用while、do…while和for语句实现(1)while#include<stdio.h>voidmain(){intsum=0,i=1;while(i<=100){sum=sum+i;i++;}printf("sum=%d\n",sum);}(2)do...while#include<stdio.h>voidmain(){intsum=0,i=1;do{sum=sum+i;i++;}while(i<=100);printf("sum=%d\n",sum);}(3)for#include<stdio.h>voidmain(){intsum=0,i;for(i=1;i<=100;i++)sum=sum+i;printf("sum=%d\n",sum);}

5、输入10个数，用冒泡法对这10个数由小到大排序。#include<stdio.h>voidmain()inta[10];inti,j,t;//数组输入/*//数组输入/*进行9次循环，实现9趟比较*//*在每趟中进行9-j次比较*//*相邻两个数进行比较*/for(i=0;i<10;i++)scanf("%d",&a[i]);printf("\n");for(j=0;j<9;j++)for(i=0;i<9-j;i++)if(a[i]>a[i+1]){t=a[i];a[i]=a[i+1];a[i+1]=t;/*满足条件进行互换*/printf("thesortednumbers:\n");for(i=0;i<10;i++) //数组输出printf("%d",a[i]);printf("\n");6、选择排序#include<stdio.h>voidmain(){inti,j,min,temp;inta[10]={6,90,45,56,1,15,44,78,58,101};//为数组赋初值//或键盘输入/*printf("enterdata:\n");£0=(1=0;1<10;1++)//输入10个数{printf("a[%d]=",i);scanf("%d",&a[i]);}printf("\n");for(i=0;i<10;i++)//输出10个数{printf("%5d",a[i]);}printf("\n");*/for(i=0;i<9;i++){min=i;for(j=i+1;j<=9;j++){if(a[min]>a[j])min=j;}temp=a[i];a[i]=a[min];a[min]=temp;}printf("\nThesortednumbers:\n");for(i=0;i<10;i++)//输出排序后的10个数{printf("%5d",a[i]);}printf("\n");getchar();}7、〃编一程序，将两个字符串连接起来。不要用函数strcat#include"stdio.h"voidmain(){chars1[80],s2[40];inti=0,j=0;printf("\ninputstring1:");scanf("%s",s1);printf("inputstring2:");scanf("%s",s2);while(s1[i]!='\0')i++;while(s2[j]!='\0')s1[i++]=s2[j++];s1[i]='\0';printf("Thenewstringis:%s\n",s1);}8、函数1：(参数为值传递)#include<stdio.h>voidmain(){intmax(intx,inty);inta，b，c；scanf("%d%d",&a,&b);c=max(a,b)；printf(“max=％d\n”，c)；}intmax(intx,inty){intz;if(x>y)z=x;elsez=y;returnz;}9、函数2：(参数为地址传递)有一维数组score，内放10个学生成绩，定义函数average求平均成绩。#include<stdio.h>voidmain(){floataverage(floatarray[10])；floatscore[10],aver;inti;printf("input10scores:\n");for(i=0;i<10;i++)scanf("%f",&score[i]);printf("\n");aver=average(score);printf("averagescoreis%5.2f\n",aver);}floataverage(floatarray[10]){inti;floataver,sum=array[0];for(i=1;i<10;i++)sum=sum+array[i];aver=sum/10;return(aver);}10、对输入的两个整数按大小顺序输出，用指针变量作函数的参数。(地址传递)#include<stdio.h>voidmain(){voidswap(int*p1,int*p2);inta,b;int*pointer_1,*pointer_2;scanf("%d%d",&a,&b);pointer_1=&a;pointer_2=&b;if(a<b)swap(pointer_1,pointer_2);/*调用函数后，a,b的值进行了交换*/printf("%d,%d\n",a,b);//等价于：printf("%d,%d\n",*pointer_1,*pointer_2);}voidswap(int*p1,int*p2){inttemp;temp=*p1;*p1=*p2;*p2=temp;}11、有一整型数组a，有10个元素。要输出各元素的值有三种方法(1)下标法#include<stdio.h>voidmain(){inta[10];inti;for(i=0;i<10;i++)scanf("%d",&a[i]);for(i=0;i<10;i++)printf("%d",a[i]);printf("\n");}(2)通过数组名计算出数组元素的地址，找出数组元素的值。#include<stdio.h>voidmain(){inta[10];inti;for(i=0;i<10;i++)scanf("%d",&a[i]);for(i=0;i<10;i++)printf("%d",*(a+i));printf("\n");}⑶用指针变量指向数组元素#include<stdio.h>voidmain(){inta[10];int*p,i;for(i=0;i<10;i++)scanf("%d",&a[i]);for(p=a;p<(a+10);p++)printf("%d",*p);printf("\n");}上述3种方法中有关数组元素的输入也可用下列方法表示：for(p=a;p<a+10;p++)scanf("%d",p);12、输入两个学生学号、姓名和成绩，输出成绩较高的学生的学号、姓名和成绩。#include<stdio.h>voidmain(){structstudent{intnum;charname[20];intscore;}s1,s2;scanf("%d%s%d",&s1.num,,&s1.score);scanf("%d%s%d",&s2.num,,&s2.score);printf("numnamescore\n");if(s1.score>=s2.score)printf("%d %s %d\n",s1.num,,s1.score);elseprintf("%d %s %d",s2.num,,s2.score);}13、有N个学生的信息(包括学号，姓名，成绩)，要求按照成绩的高低顺序存储并输出各学生的信息。#include<stdio.h>#include<string.h>voidmain(){structstudent{intnum;charname[20];intscore;}s[6]={{1,"aaa",59},{2,"bbb",78},{3,"ccc",85},{4,"ddd",64},{5,"eee",98},{6,"fff",83}};inti,j,max,temp;chartemp_name[20];for(i=0;i<6;i++){max=i;for(j=i+1;j<=5;j++){if(s[max].score<s[j].score)max=j;}/*temp_stu=s[i];s[i]=s[max];s[max]=temp_stu;*/temp=s[i].num;s[i].num=s[max].num;s[max].num=temp;strcpy(temp_name,s[i].name);strcpy(s[i].name,s[max].name);strcpy(s[max].name,temp_name);temp=s[i].score;s[i].score=s[max].score;s[max].score=temp;}printf("numnamescore\n");for(i=0;i<6;i++)printf("%d%s%d\n",s[i].num,s[i].name,s[i].score);}14、从键盘输入4个学生的有关数据，然后把它们转存到磁盘文件上去。#include<stdio.h>#defineSIZE4structstudent_type{charname[10];intnum;intage;charaddr[15];}stud[SIZE];voidmain(){voidsave();inti;for(i=0;i<SIZE;i++)/*从键盘输入*/scanf("%s%d%d%s",stud[i].name,&stud[i].num,&stud[i].age,stud[i].addr);save();}voidsave(){FILE*fp;inti;if((fp=fopen("stu-list","wb"))==NULL){printf("cannotopenfile\n");return;}for(i=0;i<SIZE;i++)if(fwrite(&stud[i],sizeof(structstudent_type),1,fp)!=1)printf(“filewriteerror\n");/*出错*/fclose(fp);/*关闭文件*/}15、将以上数据读出并显示#include<stdio.h>#defineSIZE4structstudent_type{charname[10];intnum;intage;charaddr[15];}stud[SIZE];voidmain(){inti;FILE*fp;if((fp=fopen("stu_list","rb"))==NULL){printf("cannotopenfile\n");return;}for(i=0;i<SIZE;i++){fread(&stud[i],sizeof(structstudent_type),1,fp);printf("%-10s%4d%4d%-15s\n",stud[i].name,stud[i].num,stud[i].age,stud[i].addr);}fclose(fp);}16、//有15个数按照从小到大的顺序存放在一个数组中。输入一个数，要求用折半查找法找出该数是数组中第几个元素的值。如果该数不在数组中，输出“不在表中”#include"stdio.h"#defineN15voidmain(){inti,number,top,bott,mid,loca,flag=1,sign;inta[N]={1,3,4,5,6,8,12,23,34,44,45,56,57,58,68};charc;for(i=0;i<N;i++)printf("%d",a[i]);printf("\n");while(flag){printf("inputnumbertolookfor:");scanf("%d",&number);getchar();sign=0;top=0;bott=N-1;if((number<a[0])||(number>a[N-1]))loca=-1;while((!sign)&&(top<=bott)){mid=(bott+top)/2;if(number==a[mid]){loca=mid;printf("Hasfound%d,itspositionis%d\n",number,loca+1);sign=1;elseif(number<a[mid])bott=mid-1;elsetop=mid+1;if(!sign||loca==-1)printf("cannotfind%d.\n",number);printf("continueornot(Y/N)?");scanf("%c",&c);getchar();if(c=='N'||c=='n')flag=0;10 11 12 13 1410 11 12 13 1445 56 57 58 68；45 56 57 58 6845 56 57 58 6845 56 57 58 68TOC\o"1-5"\h\z0 1 2 3 4 5