1、c=a*b;printf(%d*%d=%d,a,b,c);第3章 (1) b (2) b (3) d (4) a (5) b 2. (1)&b (2)l,s 3. printf(“x=%.2f,y=%.2fn”,x,y); int num1,num2,num3,sum; float average; scanf(%d%d%dnum1,&num2,&num3); sum=num1+num2+num3; average=sum/3.0; printf(sum=%d,average=%.2fn,sum,average); int hour,minute,second,total; /* 定义变量代表
2、时、分、秒和总秒数 */ %dtotal); hour=total/3600;minute=total%3600/60; second=total%3600%60;%dhours:%dminutes:%dsecondsn,hour,minute,second);第4章 (1) a (3) b (4) b 2. 0 (1) 6 (2) 4 (3) 1 (4) 1 int x,y;x); if (x-5 & x0) y=5*x; if (x = 0) y=-1;0 &10) y=2*x+1;x=%d,y=%dn,x,y); int score,rank; /* score 表示成绩,rank 表示
3、级别 */ Please input score:);score); rank=score/10; switch(rank) case 10: case 9:成绩等级为:An break; case 8:Bn case 7:Cn case 6:Dn default:En 6. void main() int n;Please input the digit:n); switch(n) case 1:Jan nbreak; case 2:Feb n case 3:Mar n case 4:Apr n case 5:May nJun nJul nAgu nSep nOct n case 11:No
4、v n case 12:Dec n第5章 (4) d 2. 7,1 3. y=-1 4. m=6 5. 10,14 6. 3 7. 5 8. ABABCDCD 9. # include main ( ) int i; long int sum=0; for (i=2;i=200;i+=2) sum+=i;2+4+6+.+200=%ld,sum);10. long int n; int sum = 0;Please input the nber:%ld while(n != 0) sum += n % 10; n /= 10;%dn, sum);第6章 (1) d (4) c (5) a 2.
5、11 3. 3 4. 5689 5. 12,6 6. (1) i- (2) n 7. (1) char k; (2) ij 8. float a10; float sum=0,average; for (i=0;10;i+) a%d=?,i);%fai); sum+=ai; average=sum/10;average is %.2fn,average);#define N 10 int aN,i,target,found;N;Please input a number:target); i=0; while(iN & target != ai) i+; found = ib?(ac?a:c)
6、:(bb: return max;long int sum(int n);n=?Sum=%ld.n,sum(n);long int sum(int n) if (n = = 1 ) return 1; else return sum(n-1)+n;void fun(n); fun(n);void fun(n) if (n = = 0) return; else fun(n/2);%-2d,n%2);第8章 (2) d (5) c 2. 8 3. 123456789 4. 2 3 4 5 6 5. 345 6. 1 2 3 4 5 6 7. bi 8. bcdefgha 9. p=sum str
7、ing.hchar str80;char *p1, *p2;gets(str);p1=str;p2=str+strlen(str)-1;while (p1p2 & *p1+ = *p2- ) ;puts( p1p2 ? 不是回文 :是回文 );第9章 a. #define F(x) (x)*(x)*(x) b. #define F(x) (x)%4 c. #define F(x,y) (x)*(y)typedef struct Employee int id; char name20; char gender20; int age; char address20;Employee;int ma
8、in(void) FILE *fp; char another,choice; Employee emp; long int recsize; fp=fopen(employee.datrb+ if(fp=NULL) fp=fopen( wb+Cant Open File exit(0); recsize=sizeof(emp); while(1) 1.添加记录 2.显示男员工 3.退出nEnter your choice(1-3): fflush(stdin);%cchoice); switch(choice) case1: fseek(fp,0,SEEK_END); another=Y;
9、while(another=| another=y输入信息(id 姓名 性别 年龄 住址):n%d %s %s %d %semp.id,&emp.name,&emp.gender,&emp.age,&emp.address); fwrite(&emp,recsize,1,fp);是否继续 (Y/N): another=getchar(); case 2学号t 姓名t性别t 年龄t住址n rewind(fp); while(fread(&emp,recsize,1,fp)=1) if (strcmp(emp.gender,男)=0) %dt%st%st%dt%sn emp.id,emp.name
10、,emp.gender,emp.age,emp.address);3 fclose(fp);#includestdio.hstdlib.h#define M 2 #define stu struct student stu int num; float s1; float s2; float s3; float avg; stu stM;请输入 5 名同学生的成绩,按照学号,姓名,成绩 1,成绩2,成绩 3 的顺序,中间用空格隔开: for(i=0;M;%d%s%f%f%fsti.num,sti.name,&sti.s1,&sti.s2,&sti.s3); sti.avg=(sti.s1+st
11、i.s2+sti.s3)/3; if(fp=fopen(stud.recwb)=NULL) cannot open filen if(fwrite(&sti,sizeof(stu),1,fp)!=1) file write errornrbnumtnametscore1tscore2tscore3taveragen fread(&sti,sizeof(stu),1,fp);%dt%st%.2ft%.2ft%.2ft%.2fn,sti.num,sti.name,sti.s1,sti.s2,sti.s3,sti.avg);第12章 a. 2 c. 11 d. 4 e. -15 f. 28 g. -36 程序设计分析:先将整数x 右移4 位,将该整数机内码的第4 到7 位移至第0 到3 位, 然后与0x000f(0000000000001111)进行位与运算,所保留的低4 位就是所要的结果 int x,y;y=x4; y=y&0x000f;47位的十六进制数:y=%#xn,y);4. (1) p!=NULL (2)p=p-next 5. p=p-next 6. 略
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