乘法口诀表#include <stdio.h>#include <conio.h>void main(void){int i,j,x,y;clrscr();printf("\n\n * * * 乘法口诀表* * * \n\n");x=9;y=5;for(i=1;i<=9;i++){gotoxy(x,y);printf("%2d ",i);x+=3;}x=7;y=6;for(i=1;i<=9;i++){gotoxy(x,y);printf("%2d ",i);y++;}x=9;y= 6;for(i=1;i<=9;i++){for(j=1;j<=9;j++){gotoxy(x,y);printf("%2d ",i*j);y++;}y-=9;x+=3;}printf("\n\n");}用一维数组统计学生成绩#include <stdio.h>void main(){char SelectKey,CreditMoney,DebitMoney;while(1){do{clrscr();puts("=========================");puts("| Please select key: |");puts("| 1. Quary |");puts("| 2. Credit |");puts("| 3. Debit |");puts("| 4. Return |");puts("=========================");SelectKey = getch();}while( SelectKey!='1' && SelectKey!='2' && SelectKey!='3' &&SelectKey!='4' );switch(SelectKey){case '1':clrscr();puts("================================");puts("| Your balance is $1000. |");puts("| Press any key to return... |");puts("================================");getch();break;case '2':do{clrscr();puts("==================================");puts("| Please select Credit money: |");puts("| 1. $50 |");puts("| 2. $100 |");puts("| 3. Return |");puts("==================================");CreditMoney = getch();}while( CreditMoney!='1' && CreditMoney!='2' && CreditMoney!='3' );switch(CreditMoney){case '1':clrscr();puts("=========================================");puts("| Your Credit money is $50,Thank you! |");puts("| Press any key to return... |");puts("=========================================");getch();break;case '2':clrscr();puts("==========================================");puts("| Your Credit money is $100,Thank you! |");puts("| Press any key to return... |");puts("==========================================");getch();break;case '3':break;}break;case '3':do{clrscr();puts("====================================");puts("| Please select Debit money: |");puts("| 1. $50 |");puts("| 2. $100 |");puts("| 3. $500 |");puts("| 4. $1000 |");puts("| 5. Return |");puts("====================================");DebitMoney = getch();}while( DebitMoney!='1' && DebitMoney!='2' && DebitMoney!='3' &&DebitMoney!='4' && DebitMoney!='5' );switch(DebitMoney){case '1':clrscr();puts("===========================================");puts("| Your Debit money is $50,Thank you! |");puts("| Press any key to return... |");puts("===========================================");getch();break;case '2':clrscr();puts("===========================================");puts("| Your Debit money is $100,Thank you! |");puts("| Press any key to return... |");puts("===========================================");getch();break;case '3':clrscr();puts("===========================================");puts("| Your Debit money is $500,Thank you! |");puts("| Press any key to return... |");puts("===========================================");getch();break;case '4':clrscr();puts("===========================================");puts("| Your Debit money is $1000,Thank you! |");puts("| Press any key to return... |");puts("===========================================");getch();break;case '5':break;}break;case '4':clrscr();puts("================================");puts("| Thank you for your using! |");puts("| Good bye! |");puts("================================");return;}}模拟ATM（自动柜员机）界面#include <stdio.h> void main(){int Password=0,Number=0,price=58,i=0;while( Password != 1234 ){if( i >= 3 )return;i++;puts("Please input Password: ");scanf("%d",&Password);}i=0;while( Number!=price ){do{puts("Please input a number between 1 and 100: ");scanf("%d",&Number);printf("Your input number is %d\n",Number);}while( !(Number>=1 && Number<=100) );if( Number >= 90 ){printf("Too Bigger! Press any key to try again!\n");}else if( Number >= 70 && Number < 90 ){printf("Bigger!\n");}else if( Number >= 1 && Number <= 30 ){printf("Too Small! Press any key to try again!\n");}else if( Number > 30 && Number <= 50 ){printf("Small! Press any key to try again!\n");}else{if( Number == price ){printf("OK! You are right! Bye Bye!\n");}else if( Number < price ){printf("Sorry,Only a little smaller! Press any key to try again!\n");}else if( Number > price ){printf(" Sorry, Only a little bigger! Press any key to try again!\n");}getch();}}用二维数组实现矩阵转置/* 用二维数组实现矩阵的转置*/#include <stdio.h>#define ROW 3#define COL 4main(){int matrixA[ROW][COL],matrixB[COL][ROW];int i,j; clrscr();printf("Enter elements of the matrixA,");printf("%d*%d:\n",ROW,COL);for( i=0; i<ROW; i++ ){for( j=0; j<COL; j++ ){scanf("%d",&matrixA[i][j]);}}for( i=0; i<ROW; i++ ){for( j=0; j<COL; j++ ){matrixB[j][i] = matrixA[i][j];}}printf("MatrixB,");printf("%d*%d:\n",COL,ROW);for( i=0; i<COL; i++ ){for( j=0; j<ROW; j++ ){printf("%8d",matrixB[i][j]);}printf("\n");}printf("\n Press Any Key to Quit... \n");getch();}求解二维数组的最大/最小元素#define MAXN 20int a[MAXN][MAXN];main(){int min, /* 存储最小值*/max; /* 存储最大值*/int row,col,n;clrscr();printf("Please input the order of the matrix:\n");/* 输入方阵的阶次*/ scanf("%d",&n);printf("Please input the elements of the matrix,\n from a[0][0] to a[%d][%d]:\n",n-1,n-1);for(row=0;row<n;row++)for(col=0;col<n;col++)scanf("%d",&a[row][col]);for(min=a[0][0],row=0;row<n;row++){/* 从每行选出最大数*/for(max=a[row][0],col=1;col<n;col++)/*从row行选出最大数*/if(max<a[row][col])max=a[row][col];if(min>max)/* 保存至row行的最小数*/min=max;}printf("The minimum of maximum number is %d\n",min);for(max=a[0][0],row=0;row<n;row++)/* 每行选出最小数*/for(min=a[row][0],col=1;col<n;col++)/* 从row行选出最小数*/ if(min>a[row][col])min=a[row][col];if(max<min)/*保存至row行的最大数*/max=min;}printf("The maximum of minimum numbers is %d\n",max);printf("\nPress any key to quit...\n");getch();}利用数组求前n个质数#define N 50main(){int primes[N];int pc,m,k; clrscr();printf("\n The first %d prime numbers are:\n",N);primes[0]=2;/*2是第一个质数*/pc =1;/*已有第一个质数*/m =3;/*被测试的数从3开始*/while(pc<N){/*调整m使它为下一个质数*/k=0;while(primes[k]*primes[k]<=m)if(m%primes[k]==0){/*m是合数*/m+=2;/*让m取下一个奇数*/k=1;/*不必用primes[0]=2去测试m，所以k从一开始*/}elsek++;/*继续用下一个质数去测试*/primes[pc++]=m;m+=2;/*除2外，其余质数均是奇数*/}/*输出primes[0]至primes[pc-1]*/for(k=0;k<pc;k++)printf("%4d",primes[k]);printf("\n\n Press any key to quit...\n ");getch();}编制万年历#include "stdio.h"long int f(int year,int month){/*f(年，月)＝年－1，如月<3;否则，f(年，月)＝年*/if(month<3) return year-1;else return year;} long int g(int month){/*g(月)＝月＋13，如月<3;否则，g(月)＝月＋1*/if(month<3) return month+13;else return month+1;} long int n(int year,int month,int day){/*N=1461*f(年、月)/4+153*g(月)/5+日*/return 1461L*f(year,month)/4+153L*g(month)/5+day;} int w(int year,int month,int day){/*w=(N-621049)%7(0<=w<7)*/return(int)((n(year,month,day)%7-621049L%7+7)%7);} int date[12][6][7];int day_tbl[ ][12]={{31,28,31,30,31,30,31,31,30,31,30,31},{31,29,31,30,31,30,31,31,30,31,30,31}};main(){int sw,leap,i,j,k,wd,day;int year;/*年*/char title[]="SUN MON TUE WED THU FRI SAT";clrscr();printf("Please input the year whose calendar you want to know: ");/*输入年*/scanf("%d%*c",&year);/*输入年份值和掠过值后的回车*/sw=w(year,1,1);leap=year%4==0&&year%100||year%400==0;/*判闰年*/for(i=0;i<12;i++)for(j=0;j<6;j++)for(k=0;k<7;k++)date[i][j][k]=0;/*日期表置0*/for(i=0;i<12;i++)/*一年十二个月*/for(wd=0,day=1;day<=day_tbl[leap][i];day++){/*将第i＋1月的日期填入日期表*/date[i][wd][sw]=day;sw=++sw%7;/*每星期七天，以0至6计数*/if(sw==0) wd++;/*日期表每七天一行，星期天开始新的一行*/} printf("\n|==================The Calendar of Year %d=====================|\n|",year);for(i=0;i<6;i++){/*先测算第i+1月和第i+7月的最大星期数*/for(wd=0,k=0;k<7;k++)/*日期表的第六行有日期，则wd!=0*/wd+=date[i][5][k]+date[i+6][5][k];wd=wd?6:5;printf("%2d %s %2d %s |\n|",i+1,title,i+7,title);for(j=0;j<wd;j++){printf(" ");/*输出四个空白符*//*左栏为第i+1月，右栏为第i+7月*/for(k=0;k<7;k++)if(date[i][j][k])printf("%4d",date[i][j][k]);else printf(" ");printf(" ");/*输出十个空白符*/for(k=0;k<7;k++)if(date[i+6][j][k])printf("%4d",date[i+6][j][k]);else printf(" ");printf(" |\n|");}/*scanf("%*c");/*键入回车输出下一个月的日历*/}puts("=================================================================|");puts("\n Press any key to quit...");getch();}对数组元素排序rest(int a[], int n){int i,low,high,t; for(i=0,low=0,high=n-1;i<=high;) {if(a[i]>0){/*a[i]与a[high]交换，随之high减1*/t=a[i];a[i]=a[high];a[high]=t;high--;}else if(a[i]==0)i++; /* 掠过该元素*/else{/*a[i]与a[low]交换，随之low增1, i增1*/t=a[i];a[i]=a[low];a[low]=t;low++;i++;}}int s[]={8,4,0,-1,6,0,-5};main(){int i;clrscr();printf("\n The arry before rest is:\n");for(i=0;i<sizeof(s)/sizeof(s[0]);i++)printf("%4d",s[i]);rest(s,sizeof(s)/sizeof(s[0]));printf("\n The arry after rest is:\n");for(i=0;i<sizeof(s)/sizeof(s[0]);i++)printf("%4d",s[i]);printf("\n Press any key to quit...\n");getch();}任意进制数的转换/* 函数trans将无符号整数n翻译成d（2<＝d<＝16）进制表示的字符串s */ #define M sizeof(unsigned int)*8int trans(unsigned n, int d, char s[]){static char digits[] ="0123456789ABCDEF"; /* 十六进制数字的字符*/char buf[M+1];int j, i = M;if(d<2||d>16){s[0]='\0'; /* 不合理的进制，置s为空字符串*/return 0; /* 不合理的进制，函数返回0 */}buf[i]='\0';do{buf[--i]=digits[n%d]; /*译出最低位，对应字符存入对应工作数组中*/n/=d;}while(n);/* 将译出在工作数组中的字符串复制到s */for(j=0;(s[j]=buf[i])!='\0';j++,i++);/* 其中控制条件可简写成s[j]=buf[i] */return j;}/* 主函数用于测试函数trans() */main()unsigned int num = 253;int scale[]={2,3,10,16,1};char str[33];int i;clrscr();for(i=0;i<sizeof(scale)/sizeof(scale[0]);i++){if(trans(num,scale[i],str))printf("%5d = %s(%d)\n",num,str,scale[i]);elseprintf("%5d => (%d) Error! \n",num,scale[i]);}printf("\n Press any key to quit...\n");getch();}判断回文数/* 函数circle用于判断正整数n的d进制数表示形式是否是回文数*/ int circle(int n, int d){int s=0,m=n;while(m){s=s*d+m%d;m/=d;}return s==n;}/* main函数用于测试circle函数*/int num[]={232,27,851};int scale[]={2,10,16};main(){int i,j;clrscr();for(i=0;i<sizeof(num)/sizeof(num[0]);i++)for(j=0;j<sizeof(scale)/sizeof(scale[0]);j++)if(circle(num[i],scale[j]))printf("%d -> (%d) is a Circle Number!\n",num[i],scale[j]);elseprintf("%d -> (%d) is not a Circle Number!\n",num[i],scale[j]);printf("\n Press any key to quit...\n");getch();}求解钢材切割的最佳订单#include <stdio.h>#define N 20#define DELTA 2int bestlen;int bestsele[N];int sele[N];int n;int orderlen[N];int total;main(){int i;clrscr();printf("\n Please enter total length of the steel:\n");/* 输入钢材总长*/scanf("%d",&total);printf("\n Please enter number of order:\n"); /* 输入定单数*/ scanf("%d",&n);printf("\n Please enter the orders:\n"); /* 输入各定单*/for(i=0;i<n;i++)scanf("%d",&orderlen[i]);bestlen=0; /*最佳解用料的初值*/for(i=0;i<n;i++)sele[i]=bestsele[i]=0; /*置当前选择和最佳选择初值*/try(); /* 调用函数求解*/for(i=0;i<n;i++) /* 输出结果*/if(bestsele[i])printf("order %d length = %d\n",i+1,orderlen[i]);printf("\n Press any key to quit...");getch();}try(){int i,len;for(len=i=0;i<n;i++) /* 求当前选中的用料量*/if(sele[i])len+=orderlen[i]+DELTA;if(len-DELTA<=total) /* 注意最后一段可能不需要切割*/{if(bestlen < len){/* 找到一个更好的解*/bestlen = len;for(i=0;i<n;i++)bestsele[i]=sele[i];}for(i=0;i<n;i++) /* 对所有未选定单逐一作选中尝试循环*/if(!sele[i]){sele[i]=1; /* 做选中尝试*/try();sele[i]=0;}}}指向数组的指针main(){int x,y,z; /* 定义三个int型变量*/int *xp = &x, /* 定义指针变量xp,并赋值为x的地址，使xp指向x */ *yp = &y, /* 定义指针变量yp,并赋值为y的地址，使yp指向y */*zp = &z; /* 定义指针变量zp,并赋值为z的地址，使zp指向z */int t;clrscr();printf("\nPlease input x,y,z:\n");scanf("%d%d%d",xp,yp,zp); /* 通过变量的指针，为变量输入值*/ if(*xp>*yp) /* 通过指向变量的指针引用变量的值*/{t=*xp; /* 通过指向变量的指针引用变量的值*/*xp=*yp;/* 通过指向变量x的指针xp,引用变量x的值*/*yp=t; /* 通过指向变量y的指针yp,引用变量y的值*/}if(*xp>*zp) /* 通过指向变量的指针,引用变量的值*/{t=*xp; /* 通过指向变量x的指针xp,引用变量x的值*/*xp=*zp;/* 通过指向变量x的指针xp,引用变量x的值*/*zp=t; /* 通过指向变量z的指针zp,引用变量z的值*/}if(*yp>*zp) /* 通过指向变量的指针,引用变量的值*/{t=*yp; /* 通过指向变量的指针,引用变量的值*/*yp=*zp;/* 通过指向变量y的指针yp,引用变量y的值*/*zp=t;/* 通过指向变量z的指针zp,引用变量z的值*/}printf("x = %d\ty = %d\tz = %d\n",x,y,z);printf("\nPress any key to quit...\n");getch();}阿拉伯数字转换为罗马数字#include <stdio.h>#define ROWS 4#define COLS 4int nums[ROWS][COLS]={{1000,1000,1000,1000},{900,500,400,100},{90,50,40,10},{9,5,4,1}};char *roms[ROWS][COLS]={{"m","m","m","m"}, {"cm","d","cd","c"},{"xc","l","xl","x"},{"ix","v","iv","i"}}; main(int argc,char *argv[ ]){int low,high;char roman[25]; if(argc<2){ printf("Usage:roman decimal_number\n");/*运行程序需带整数参数*/ exit(0);}high=low=atoi(argv[1]);/*将第一个参数转换成整数*/checknum(low);if(argc>2){/*带两个参数*/high=atoi(argv[2]);checknum(high);if(low>high){low=high;high=atoi(argv[1]);}}elselow=1;for(;low<=high;low++){to_roman(low,roman);printf("%d\t%s\n",low,roman);}} checknum(int val)/*检查参数合理性*/{if(val<1||val>9999){printf("The number must be in range 1..9999.\n");exit(0);}}to_roman(int decimal,char roman[ ])/*将整数转换成罗马数字表示*/ {int power,index;roman[0]='\0';for(power=0;power<ROWS;power++)for(index=0;index<COLS;index++)while(decimal>=nums[power][index]){strcat(roman,roms[power][index]);decimal-=nums[power][index];}}通讯录的输入输出#include <stdio.h>#define ZIPLEN 10#define PHONLEN 15/*struct addr类型定义*/ struct addr{char *name;/*姓名*/char *address;/*地址*/char zip[ZIPLEN];/*邮政编码*/char phone[PHONLEN];/*电话号码*/}; main()/*本主函数示意上述输入输出函数的用法*/{struct addr p[100];int i,j;clrscr();for(i=0;readaddr(p+i);i++);for(j=0;j<i;j++) writeaddr(p+j);puts("\n Press any key to quit...");getch();} /* 函数readaddr用于输入一个通信录函数*/int readaddr(struct addr *dpt){int len;char buf[120];/*输入字符串的缓冲区*/ printf("\nPlease input theName:\n");/*输入姓名*/if(scanf("%s",buf)==1){len=strlen(buf);dpt->name=(char *)malloc(len+1);/*申请存贮姓名的空间*/ strcpy(dpt->name,buf);}else return 0;/*Ctrl+Z结束输入*/printf("Please input the Address:\n");/*输入地址*/if(scanf("%s",buf)==1){len=strlen(buf);dpt->address=(char *)malloc(len+1);/*申请存贮地址的空间*/ strcpy(dpt->address,buf);}else{/*Ctrl+Z结束输入*/free(dpt->name);/*释放存贮姓名的空间*/return 0;printf("Please input the Zip code:\n");/*输入邮编*/if(scanf("%s",buf)==1)strncpy(dpt->zip,buf,ZIPLEN-1);else{free(dpt->name);/*释放存贮姓名的空间*/free(dpt->address);/*释放存贮地址的空间*/return 0;/*Ctrl+Z结束输入*/}printf("Please input the Phone number:\n");/*输入电话号码*/if(scanf("%s",buf)==1)strncpy(dpt->phone,buf,PHONLEN-1);else{free(dpt->name);free(dpt->address);return 0;/*Ctrl+Z结束输入*/}return 1;} /* 函数writeaddr用于输出通讯录*/int writeaddr(struct addr*dpt){printf("Name : %s\n", dpt->name);/*输出姓名*/printf("Address : %s\n", dpt->address);/*输出地址*/printf("Zip : %s\n", dpt->zip);/*输出邮编*/printf("Phone : %s\n\n", dpt->phone);/*输出电话号码*/}扑克牌的结构表示enum suits{CLUBS,DIAMONDS,HEARTS,SPADES};struct card{enum suits suit;char value[3];};struct card deck[52];char cardval[][3]={"A","2","3","4","5","6","7","8","9","10","J","Q","K"};char suitsname[][9]={"CLUBS","DIAMONDS","HEARTS","SPADES"}; main()int i,j;enum suits s;clrscr();for(i=0;i<=12;i++)for(s=CLUBS;s<=SPADES;s++){j=i*4+s;deck[j].suit=s;strcpy(deck[j].value,cardval[i]);}for(j=0;j<52;j++)printf("(%s%3s)%c",suitsname[deck[j].suit],deck[j].value,j%4==3?'\n':'\t');puts("\nPress any key to quit...");getch();}用“结构”统计学生成绩#include <stdio.h>#define N 200#define SCORES 5#define NUMLEN 10struct std_type{char no[NUMLEN];/*学号*/char *name;/*名字符串指针*/int scores[SCORES];/*五门功课的成绩*/};struct std_type students[N];int order[N];int total[N]; /*[函数]输入一个学生信息函数*/int readastu(struct std_type *spt){int len,j;char buf[120];/*输入字符串的缓冲区*/ printf("\nNumber : ");/*输入学号*/if(scanf("%s",buf)==1)strncpy(spt->no,buf,NUMLEN-1);elsereturn 0;/*Ctrl+Z结束输入*/printf("Name : ");/*输入姓名*/if(scanf("%s",buf)==1){len=strlen(buf);spt->name=(char *)malloc(len+1);/*申请存贮姓名的空间*/ strcpy(spt->name,buf);}else return 0;/*Ctrl+Z结束输入*/printf("Scores : ");/*输入成绩*/for(j=0;j<SCORES;j++)if(scanf("%d",spt->scores+j)!=1)break;if(j==0)/*一个成绩也未输入*/{free(spt->name);/*释放存贮姓名的空间*/return 0;}for(;j<SCORES;j++)/*少数未输入的成绩用0分代之*/ spt->scores[j]=0;return 1;} /*[函数]输出一个学生信息的函数*/int writeastu(struct std_type *spt){int i; printf("Number : %s\n",spt->no);/*输出学号*/printf("Name : %s\n",spt->name);/*输出姓名*/printf("Scores : ");/*输出成绩*/for(i=0;i<SCORES;i++)printf("%4d",spt->scores[i]);printf("\n\n");} main(){int n,i,j,t; clrscr();for(n=0;readastu(students+n);n++);/*采用冒泡法对学生信息数组排序*/for(i=0;i<n;i++){order[i]=i;/*预置第i个输入的学生*/for(t=0,j=0;j<SCORES;j++)/*求第i个学生的总分*/t+=students[i].scores[j];total[i]=t;}/*冒泡排序*/for(i=0;i<n-1;i++)/*共扫视n-1遍*/for(j=0;j<n-1-i;j++)if(total[order[j]]<total[order[j+1]]){/*交换名次*/t=order[j];order[j]=order[j+1];order[j+1]=t;}for(j=0;j<n;j++)/*输出*/writeastu(students+order[j]);printf("\n Press any key to quit...\n");getch();}报数游戏#include <stdio.h>struct ele{int no;struct ele *link;}main(){int n,m,i;struct ele *h,*u,*p;clrscr();printf("Please input n&m:\n");scanf("%d%d",&n,&m);/*输入n和m*/h=u=(struct ele *)malloc(sizeof(struct ele));/*形成首表元*/ h->no=1;for(i=2;i<=n;i++)/*形成其余的n-1个表元*/{u->link=(struct ele *)malloc(sizeof(struct ele));u=u->link;u->no=i;/*第i个表元置编号i*/}u->link=h;/*末表元后继首表元，形成环*/puts("\nThe numbers of who will quit the cycle in turn are:"); while(n){for(i=1;i<m;i++)/*掠过m－1个表元*/u=u->link;p=u->link;/*p指向第m个表元*/u->link=p->link;/*第m个表元从环中脱钩*/printf("%4d",p->no);free(p);/*释放第m个表元占用的空间*/n--;}printf("\n\n Press any key to quit...\n");getch();}学生成绩管理程序/*学生成绩管理程序编制一个统计学生考试分数的管理程序。
