1、 a=513.789185,a=513.79,a=513.78918457,a=513.78918457(2) a=3b=7x=8.5y=71.82c1=Ac2=a3编程题(1) int x,y;%d%dx,&y);商数=%d,余数=%d,x/y,x%y); double x,y,z,avg;%lf%lf%lfy,&z); avg=(x+y+z)/3;%.1f,avg);第4章 (1)(10) CCAAD CCABD (1)a0&b0|ac0|bab|a5|-a5(a0) 1 !=0&x=AchZ ch=ch-32(5) x2 xc&a+cb&b+ca a=b&a=c a=b|a=c|b=c(
2、7) x-5 & x0 &10) y=x+1;the number is error(3) int a,m;a); switch(a/10) case 0: case 1: case 2:m=1;break; case 3:m=2; case 4:m=3; case 5:m=4; default:m=5;,m);(4) float price,tax;please input the price of product:price); if(price=10000) tax=price*0.05; else if(price=5000) tax=price*0.03;=1000) tax=pri
3、ce*0.02; else tax=0;,tax);(5) float score;please input the score of student:score); if(score=85) printf(VERY GOOD else if(score=60) printf(GOODBAD(6)# include main () int x,y,z;input two number: if(x99 | y99)input data error!n else z=(x/10)*1000+(y/10)*100+(x%10)*10+y%10;%dn,z);第5章(1)d (2) c (3)a (4
4、)d (5)a (6)d (7)d (8)b (9)d (10)b(11)c (12)a (13)d (14)d (15)c(1) fahr celsius=5.0/9*(fahr-32) fahr+=step (2) 5、4、6 (3) 3*i-2(4) y- z*=x (5) 852 (6) j+ i%j=0 j=i(或j=i) (7)sumk sum=k j-2 (8) s=0 p=1 j=i 3改错题 (1) 第一处改正: For改为for 第二处改正:for(k=1;k=eps第三处改正:r*2(4) 第一处改正:n= =(5) 第一处改正:t=m/10改为t=m%10t=0改为t=
5、0m=m%10改为m=/104编程题(1) 分式累加和。math.h int s; float n,t,sum; t=1; sum=0; n=1; s=1.0; while(n=100) sum=sum+t; n=n+1; s=-s; t=s/n;sum=%10.6fn,sum);(2)110的阶乘 int i; long int n=1; for(i=1;i=10;i+) n=n*i; printf(i%5=0?%2d!=%-10ldn:=%-10ld,i,n);(3)由*组成的等腰三角形 int n,i,j;n);=n; for(j=1;j=n+i-1;j+) if(j=n-i) prin
6、tf( *(4)完数 int m,s,i; for(m=2;m1000;m+) s=0;m; if(m%i)=0) s=s+i; if(s=m) printf(%d its factors are if(m%i=0) printf(%d,i);(5)打靶 int h10,h7,h5; for(h10=0; h10=8; h10+) for(h7=0; h7 h7+) for(h5=0; h50) x1=(x2+1)*2; /*第一天的桃子数是第2天桃子数加1后的2倍*/ x2=x1;day-; printf(thetotalis,x1);/代码二 int i=9,sum=1; for(; i0
7、; sum=2*(sum+1),i-); printf( sum=%dn (7)牛顿迭代法一般地:f(x)=a0xn+a1xn-1+.+an-1x+an=0 求f(x)在X0附近的根 计算公式:Xn+1=Xn-f(Xn)/f(Xn) 精度:= |Xn+1-Xn|1.0e-m,m=5。所求的根:满足精度的Xn#include=1e-5);The root is %8.7fn,x);(8) 利用辗除法,直到b为0为止 int p,r,n,m,temp;input two integer n,m:%d,%dn,&m); if(nm) temp=n; n=m; m=temp; p=n*m; while
8、(m!=0) r=n%m; m=r;greatest common divisor is:,n); /最大公约数lease common multiple is:,p/n); /最小公倍数第6章12 3 4 5 67 8 90CCDCC AADBD2. 写出程序的运行结果(1) sum=6 (2) Max=9,row=1,col=0 (3) 1 02 1 03 2 1 0(4) -53、填空(1)i=j=k=0 i4&j=5 j0 n%10+0x30或n%10+0 sj=0(4)si 、 s0,s1 、str,s1、s2,str4.改错题n=strlen(str)strj= strj!(2)第
9、一处改正:ch=stri(j=0) & (chtime.h int a12,count=0,i; srand(unsigned) time(NULL); for (i=0;12; ai=5+rand()%11; for(i=0;%5d,ai); count+; if(count%4=0)printf() ;#define N 5 int aN,max,min,i,sum=0,count=0; float av; imax)max=ai; if(ai=90) count90+; count80+; count70+; count60+; else if(scorei=80)=70)=60)059
10、:%dn6069:%dn7079:%dn8089:%dn90100:,30-count60,count60-count70,count70-count80,count80-count90,count90);n60 or higher:%dn70 or higher:%dn80 or higher:%dn90 or higher:,count60,count70,count80,count90);1、 char a45; int i,j,k;45;i+)ai=;5; for(j=0;i; a9*i+j= for(k=0; a9*i+j+k=* j=0;%c j+; if(j%9=0)printf
11、(2、 char a59;9; aij= aij+k= printf(,aij); (5)# define N 4 int i,j,aN=21,12,13,24,25,16,47,38,29,11,32,54,42,21,33,10;nThe original array: for(j=0;%2d aij=aij+aji; aji=0;nThe result is:第7章12345 67890 12345DADBD CADAC BDDBD 2. 填空(1) 函数 (2) main函数 (3)函数首部、函数体 (4) 形式参数、实际参数、实参、形参 (5)常量 、 变量、表达式 (6)函数定义时函数值、返回值 (7)max(max(a,b),c)、 (xy?x:y) (8)*#(9) str =abdef (10) low=high (high+low)/23.改错题(1)第一处:void改为int第二处:=n & (n%i=0)改为i (n%i!=0)或in &第三处:n!=i改为n=i(2)第一处:删除; for(j=1; j=1e-6)改为if(fabs(x1-x0)=1e-6)4 、编程题(1) long act(long n) if (n=1|n=0) return 1; return n*act(n-1);main
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