max=a[i];
else if(min>a[i])
min=a[i];
}
average=sum/20;
printf("\nsum=%d, max=%d, min=%d, average=%d \n",sum,max,min,average);
puts("\nany key to exit!");
getche();
}
5、编程在一个已知的字符串中查找最长单词,假定字符串中只含字母和空格,空格用来分隔不同单词。
#include
#include
void main()
{
char string[80],*p;
int i=0,max=0;
clrscr();
printf("please input a string:\n");
gets(string);
// printf("\n%s\n",string);
p=string;
while(*p!='\0')
{
if(*p==' ')
{
if(max<=i)
max=i;
i=0;
}
else
i++;
p++;
}
if(max<=i)
max=i;
printf("\nmax_length of the string is: %d \n",max);
getche();
}
7、模拟n个人参加选举的过程,并输出选举结果:假设候选人有四人,分别用A、B、C、
D表示,当选某候选人时直接输入其编号(编号由计算机随机产生),若输入的不是A、B、
C、D则视为无效票,选举结束后按得票数从高到低输出候选人编号和所得票数。
#include
#include
#include
#include
void main()
{
int i,n,A,B,C,D,abandon,ran; //abandon stand for the ones who abandon
clrscr();
A=B=C=D=abandon=0;
printf("Enter the number of persons:");
scanf("%d",&n); //n persons to select from candidate a,b,c and d.
randomize();
for(i=1;i<=n;i++)
{
ran=random(5)+65; //A,B,C,D and the ones who abandon
switch(ran)
{
case 'A':A++;break;
case 'B':B++;break;
case 'C':C++;break;
case 'D':D++;break;
default :abandon++;
}
}
printf("\nA=%d,B=%d,C=%d,D=%d,abandon=%d\n\n",A,B,C,D,abandon);
i=4;
while(i--!=0)
{
if(A>=B&&A>=C&&A>=D&&A!=-1)
{
printf("A=%d,",A);
A=-1;
}
if(B>=A&&B>=C&&B>=D&&B!=-1)
{
printf("B=%d,",B);
B=-1;
}
if(C>=A&&C>=B&&C>=D&&C!=-1)
{
printf("C=%d,",C);
C=-1;
}
if(D>=A&&D>=B&&D>=C&&D!=-1)
{
printf("D=%d,",D);
D=-1;
}
}
printf("abandon=%d, \n\n",abandon);
getch();
}
7.任何一个自然数m的立方均可写成m个连续奇数之和。例如:1^3=1
2^3=3+5
3^3=7+9+11
4^3=13+15+17+19
编程实现:输入一自然数n,求组成n3的n个连续奇数。
#include
#include
#include //
void main()
{
int i,j,n,sum,count;
clrscr();
printf("\n an integer here please:");
scanf("%d",&n);
printf("\n");
// i=(int)floor(sqrt((float)n));
// if(i%2==0)
// i++;
i=1; //
for(;i<=n*n*n;i+=2)
{
sum=0;
count=0;
for(j=i;j<=n*n*n;j+=2)
{
sum=sum+j;
count++;
if((n*n*n==sum)&&(count==n))
break;
}
if((sum==n*n*n)&&(count==n))
break;
}
printf("i=%d,j=%d\n\n",i,j);
printf("%d*%d*%d=",n,n,n);
// while(n-->1) { printf("%d+",i); i+=2; }
for(;iprintf("%d+",i); //
printf("%d",i);
getch();
}
8、已知abc+cba=1333,其中a,b,c均为一位数,编程求出满足条件的a,b,c所有组合。
#include
#include
void main()
{
int num,a,b,c;
clrscr();
printf("Press any key to calculate!\n\n");
getch();
for(num=101;num<1000;num++)
{
a=num/100;
b=num%100/10;
c=num%10;
if(num+c*100+b*10+a==1333)
printf("\n%d+%d=1333\n",num,c*100+b*10+a);
}
while(!kbhit()) ;
}
8、编制一个完成两个数的四则运算程序。如:用户输入34+56则输出结果90.00。要求运
算结果保留两位小数,用户输入时一次将两个数和操作符输入。
#include
#include
void main()
{
float num1,num2;
char s;
clrscr();
printf("Enter a statement please:");
scanf("%f%c%f",&num1,&s,&num2);
switch(s)
{
case '+':printf("%.2f%c%.2f=%.2f",num1,s,num2,num1+num2);break;
case '-':printf("%.2f%c%.2f=%.2f",num1,s,num2,num1-num2);break;
case '*':printf("%.2f%c%.2f=%.2f",num1,s,num2,num1*num2);break;
case '/':printf("%.2f%c%.2f=%.2f",num1,s,num2,num1/num2);break;
default: printf("Input Error!");
}
getch();
}
9、输入一个五位以内的正整数,(1)判断它是一个几位数;(2)请按序输出其各位数字;
(3)逆序输出其各位数字。
如输入:56439,输出:5位数
5,6,4,3,9
9,3,4,6,5
#include
#include
long power(int n)
{
int i;
long result=1;
for(i=1;i<=n;i++)
result=10*result;
return result;
}
void main()
{
long num,n,j;
int count=1,tmpcount;
clrscr();
printf("Please input a number(0-99999):");
scanf("%ld",&num);
n=num;
while((n=n/10)!=0)
count++;
printf("\n%ld is a %d digits.\n\n",num,count);
tmpcount=count; //temporary tmpcount for later use in statement A
n=num;
while(count-->1)
{
j=power(count);
printf("%d, ",n/j);
n=n%j;
}
printf("%d\n",n);
n=num;
while(tmpcount-->1) // statement A
{
printf("%d, ",n%10);
n=n/10;
}
printf("%d\n",n);
getch();
}
10、编写子函数:(1)用冒泡法将一个数组排成升序的函数---SUB1;(2)在升序数
组中插入一个数,并且保持该数组仍为升序数组的函数---SUB2。主函数:①输入任意10个正整数给数组;②调用SUB1对数组进行排序;③从键盘输入一个正整数,调用SUB2将其插入该数组。
#include
#include
void main()
{
int i,k,a[12]={0}; //a[0] for no use
void sub1(int b[]),sub2(int b[],int k);
clrscr();
printf("Please input 10 numbers:");
for(i=1;i<=10;i++)
scanf("%d",&a[i]);
getchar();
sub1(a);
for(i=1;i<=10;i++)
printf("\na[%d]=%d\n",i,a[i]);
printf("\n\nplease input a number to be inserted into the array:");
scanf("%d",&k);
sub2(a,k);
for(i=1;i<=11;i++)
printf("\na[%d]=%d\n",i,a[i]);
puts("\nAny key to exit!");
getch();
}
void sub1(b)
int b[];
{
for (i=1;i<10;i++) //the first one is always the smallest
for(j=i;j<=10;j++)
if (b[i]>b[j])
{
t=b[i];
b[i]=b[j];
b[j]=t;
}
}
void sub2(int b[],int k)
{
int i;
for(i=10;i>=1;i--)
{
if(k
b[i+1]=b[i];
else
{
b[i+1]=k;
break;
}
}
}
11、编写函数:(1)用选择法将数组排成降序的函数----SUB1;(2)用折半查找法查
找某数是否在给定的数组当中的函数----SUB2。主函数:输入任意10个正整数给数组,调用SUB1对数组进行排序,从键盘输入一个正整数,调用SUB2在数组中进行查找,找到后输出“OK”,没有找到则输出“NO FOUND!”。
#include
#include
void main()
{
int i,key,a[11]={0},sub1(),sub2();
printf("please input 10 number: ");
for(i=1;i<=10;i++)
scanf("%d",&a[i]);
getchar();
sub1(a);
for(i=0;i<=10;i++)
printf("a[%d]=%d ,",i,a[i]);
printf("\n please input a key number: ");
scanf("%d",&key);
sub2(a,key,1,10);
}
int sub1(int b[])
{
int t,i,j,post;
for (i=1;i<10;i++)
{
post=i;
for(j=i+1;j<=10;j++)
if (b[post]>b[j])
post=j;
if(post!=i)
{
t=b[i];
b[i]=b[post];
b[post]=t;}
}
return 0;
}
int sub2(int c[],int k,int n0,int n1)
{
int i=n0,j=n1,m;
m=(i+j)/2;
while(i<=j)
{
if(kj=m-1;
if(k>c[m])
i=m+1;
if(k==c[m])
break;
m=(i+j)/2;
}
if(k==c[m])
printf("OK!\n") ;
else
printf("NO FOUND!\n");
return 0;
}
12、编写一个程序,输入两个包含5个元素的数组,先将两个数组升序排列,然后将这
两个数组合并成一个升序数组。
#include
#include
void main()
{
int i,j,k,a[6]={0},b[6]={0},c[11]={0},sub1(); clrscr();
printf("\nplease input 5 int numbers to array1: "); for(i=1;i<=5;i++) //a[0] for no use
scanf("%d",&a[i]);
getchar();
sub1(a,5);
printf("\nplease input 5 int numbers to array2: "); for(i=1;i<=5;i++) //b[0] for no use
scanf("%d",&b[i]);
getchar();
sub1(b,5);
printf("\nthe sorted array a is:\n\n");
for(i=1;i<=5;i++)
printf("a[%d]=%d ",i,a[i]);
printf("\n");
printf("\nthe sorted array b is:\n\n");
for(i=1;i<=5;i++)
printf("b[%d]=%d ",i,b[i]);
k=i=j=1;
while(i<=5&&j<=5)
if(a[i]
c[k++]=a[i++]; //c[0] for no use
else
c[k++]=b[j++];
if(ifor(;i<=5;i++)
c[k++]=a[i];
else //appending the rest ones in array b
for(;j<=5;j++)
c[k++]=b[j];
printf("\n\n");
printf("\nthe merged array c is:\n\n");
for(k=1;k<=10;k++)
{
if(k==6)
printf("\n");
printf("c[%d]=%d ",k,c[k]);
}
while(!kbhit());
}
int sub1(int b[],int n)
{
int t,i,j,post;
for(i=1;i{
post=i;
for(j=i+1;j<=n;j++)
if(b[post]>b[j])
post=j;
if(post!=i) j
{ j
t=b[i];
b[i]=b[post];
b[post]=t;
}
}
return 0;
}
13、耶稣有13个门徒,其中有一个就是出卖耶稣的叛徒,请用排除法找出这位叛徒:
13人围坐一圈,从第一个开始报号:1,2,3,1,2,3……,凡是报到“3”就退出圈子,最后留在圈内的人就是出卖耶稣的叛徒,请找出它原来的序号。
/*
// approach one
#define N 13
#include
#include
struct person
{
int number; //its order in the original circle
int nextp; //record its next person
};
struct person link[N+1]; //link[0] for no use
void main()
{
int i,count,next; //count for 12 persons,and
//next for the person not out of circle yet
clrscr();
for (i=1;i<=N;i++)
{
link[i].number=i; //numbering each person
if(i==N)
link[i].nextp=1;
else
link[i].nextp=i+1; //numbering each next person
}
printf("\nThe sequence out of the circle is:\n");
for(next=1,count=1;counti=1;
while (i!=3) //i counts 1,2,3
{
do //skip the ones whose numbers are zero
next=link[next].nextp;
while(link[next].number==0); //end of do
i++;
}
printf("%3d ",link[next].number);
link[next].number=0; //indicate out of circle already
do //start from the ones whose numbers are not zero next time next=link[next].nextp;
while(link[next].number==0);
}
printf("\n\nThe betrayer of them is:");
for(i=1;i<=N;i++)
if(link[i].number)
printf("%3d\n",link[i].number);
getch();
}
*/
// approach two using cyclic list
#define N 13
#define LEN sizeof(struct person)
#include
#include
#include
#include
// struct person //permit struct placed here//
// {
// int number;
// struct person *next;
// };
void main()
{
int i,count;
struct person //or permit struct placed here also//
{
int number;
struct person *next;
};
struct person *head,*p1,*p2;
clrscr();
head=p2=NULL;
for(i=1;i<=N;i++)
{
p1=(struct person *)malloc(LEN);
p1->number=i;
if(head==NULL)
head=p1;
else
p2->next=p1;
p2=p1;
}
p2->next=head;
printf("\nthe sequence out of the circle is:\n");
for (count=1;count{
i=1;
while(i!=3)
{
p1=head;
head=head->next;
i++;
}
p2=head;
printf("%3d ",p2->number);
p1->next=head=p2->next;
free(p2);
}
printf("\nThe betrayer of them is:\n%3d",head->number);
getch();
}
15、按如下图形打印杨辉三角形的前10行。其特点是两个腰上的数都为1,其它位置上的每一个数是它上一行相邻两个整数之和。
1
2 1
2 2 1
2 3 3 1
2 4 6 4 1
……
#include
#include
#define N 10
void main()
{
int i,j,k,a[N][N];
clrscr();
for(i=0;i{
a[i][0]=1;
a[i][i]=1;
}
for(i=2;ifor(j=1;j
a[i][j]=a[i-1][j-1]+a[i-1][j];
for(i=0;i{
for(k=0;k<=3*(N-i);k++)
printf(" ");
for(j=0;j<=i;j++)
printf("%6d",a[i][j]);
printf("\n\n");
}
getch();
}
16、某班有5个学生,三门课。分别编写3个函数实现以下要求:
(1)求各门课的平均分;
(2)找出有两门以上不及格的学生,并输出其学号和不及格课程的成绩;(3)找出三门课平均成绩在85-90分的学生,并输出其学号和姓名
主程序输入5个学生的成绩,然后调用上述函数输出结果。
#define SNUM 5 /*student number*/
#define CNUM 3 /*course number*/
#include
#include
/*disp student info*/
void DispScore(char num[][6],char name[][20],float score[][CNUM])
{
int i,j;
printf("\n\nStudent Info and Score:\n");
for(i=0;i{
printf("%s ",num[i]);
printf("%s ",name[i]);
for(j=0;jprintf("%8.2f",score[i][j]);
printf("\n\n");
}
}
/*calculate all student average score*/
void CalAver(float score[][CNUM])
{
float sum,aver;
int i,j;
for(i=0;i{
sum=0;
for(j=0;jsum=sum+score[j][i];
aver=sum/SNUM;
printf("Average score of course %d is %8.2f\n",i+1,aver); }
}
/*Find student: two courses no pass*/
void FindNoPass(char num[][6],float score[][CNUM]) {
int i,j,n;
printf("\nTwo Course No Pass Students:\n");
for(i=0;i{
n=0;
for(j=0;jif(score[i][j]<60)
n++;
if(n>=2)
{
printf("%s ",num[i]);
for(j=0;jif(score[i][j]<60)
printf("%8.2f",score[i][j]);
printf("\n");
}
}
}
/*Find student: three courses 85-90*/
void FindGoodStud(char num[][6],char name[][20],float score[][CNUM])
{
int i,j,n;
printf("\nScore of three courses between 85 and 90:\n");
for(i=0;i{
n=0;
for(j=0;jif(score[i][j]>=85&&score[i][j]<=90)
n++;
if(n==3)
printf("%s %s\n",num[i],name[i]);
}
}
/*input student info*/
void main()
{
char num[SNUM][6],name[SNUM][20]; //array num refers to student number
float score[SNUM][CNUM]; //and its length is 6
int i,j;
clrscr();
printf("\nPlease input student num and score:\n");
for(i=0;i
{
printf("\n\nStudent%d number: ",i+1);
scanf("%s",num[i]);
printf("\nStudent%d name: ",i+1);
scanf("%s",name[i]);
printf("\nStudent%d three scores: ",i+1);
for(j=0;jscanf("%f",&score[i][j]);
}
DispScore(num,name,score);
CalAver(score);
FindNoPass(num,score);
FindGoodStud(num,name,score);
getch(); }
17、编写一人个求X的Y次幂的递归函数,X为double型,y为int型,要求从主函数输入x,y的值,调用函数求其幂。
#include
#include
double fact(double x,int y)
{
if(y==1)
return x;
else
return x*fact(x,y-1);
}
void main()
{
double x;
int y;
clrscr();
printf("\nPlease x,y:");
scanf("%lf%d",&x,&y);
printf("\nx^y=%.2lf",fact(x,y));
getch();
}
18、打印魔方阵。
所谓魔方阵是指这样的的方阵:
它的每一行、每一列和对角线之和均相等。
输入n,要求打印由自然数1到n2的自然数构成的魔方阵(n为奇数)。
例如,当n=3时,魔方阵为:
8 1 6
3 5 7
4 9 2
魔方阵中各数排列规律为:
①将“1”放在第一行的中间一列;
②从“2”开始直到n×n为止的各数依次按下列规则存放:每一个数存放的行比前一个数的行数减1,列数同样加1;
③如果上一数的行数为1,则下一个数的行数为n(最下一行),如在3×3 方阵中,1在第1行,则2应放在第3行第3列。
④当上一个数的列数为n时,下一个数的列数应为1,行数减1。如2在第3行第3列,3应在第2行第1列。
⑤如果按上面规则确定的位置上已有数,或上一个数是第1行第n列时,则把下一个数放在上一个数的下面。如按上面的规定,4应放在第1行第2列,但该位置已被1占据,所以4就放在3的下面。由于6是第1行第3列(即最后一列),故7放在6下面。
#include
#include
#define Max 15
void main()
{
int i,row,col,odd;
int m[Max][Max];