1、long double:,sizeof(long double);char:,sizeof(char);(2)#define RAT 1.60934 float k;input the km:); scanf(%f,&k);mile:%fn,k/RAT);第3章 12 345 67 890(1) (10):DDCDD DCDCC 2解析题(1) x=170,x=170,x=252,x=aa,x=170 x=170,x=170,x=170,x=%6d a=513.789185,a=513.79,a=513.78918457,a=513.78918457(2) a=3b=7x=8.5y=71.82
2、c1=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(7) x-5 & x0 &10) y=x+1;the number is error(3) int a,m;a); switch(a/10) cas
3、e 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=price*0.02; else tax=0;,tax);(5) float score;please input the score of studen
4、t: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) printf(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)d (5)a (6)d (7)d (8)b (9)d (10)b(11)c (12)a (13)d (14)d (15)c(1)
5、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=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=
6、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) printf( *(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; sum=2*(sum+1),i-); printf( sum=%dn (7)牛顿迭代法一般地:f
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