C第三章課后習(xí)題答案譚浩強(qiáng)

1、.3.2題#include #include using namespace std;int main ( )float h,r,l,s,sq,vq,vz; const float pi=3.1415926; coutrh; l=2*pi*r; s=r*r*pi; sq=4*pi*r*r; vq=3.0/4.0*pi*r*r*r; vz=pi*r*r*h; coutsetiosflags(ios:fixed)setiosflags(ios:right) setprecision(2); coutl= setw(10)lendl; couts= setw(10)sendl; coutsq=set

2、w(10)sqendl; coutvq=setw(10)vqendl; coutvz=setw(10)vzendl; return 0; 3.3題#include using namespace std;int main ()float c,f; coutf; c=(5.0/9.0)*(f-32); /注意5和9要用實(shí)型表示,否則5/9值為0 cout攝氏溫度為:cendl; return 0;3.4題#include using namespace std;int main ( )char c1,c2; cout請(qǐng)輸入兩個(gè)字符c1,c2:; c1=getchar(); /將輸入的第一個(gè)字符賦

3、給c1 c2=getchar(); /將輸入的第二個(gè)字符賦給c2 cout用putchar函數(shù)輸出結(jié)果為:;精品. putchar(c1); putchar(c2); coutendl; cout用cout語(yǔ)句輸出結(jié)果為:; coutc1c2endl; return 0;3.4題另一解#include using namespace std;int main ( )char c1,c2; cout請(qǐng)輸入兩個(gè)字符c1,c2:; c1=getchar(); /將輸入的第一個(gè)字符賦給c1 c2=getchar(); /將輸入的第二個(gè)字符賦給c2 cout用putchar函數(shù)輸出結(jié)果為:; putch

4、ar(c1); putchar(44); putchar(c2); coutendl; cout用cout語(yǔ)句輸出結(jié)果為:; coutc1,c2endl; return 0;3.5題#include using namespace std;int main ( )char c1,c2; int i1,i2; /定義為整型 couti1i2; c1=i1; c2=i2; cout按字符輸出結(jié)果為:c1 , c2endl; return 0;3.8題#include using namespace std;int main ( ) int a=3,b=4,c=5,x,y; coutc & b=c)

5、endl; cout(a|b+c & b-c)endl;精品. coutb) & !c|1)endl; cout(!(x=a) & (y=b) & 0)endl; cout(!(a+b)+c-1 & b+c/2)endl; return 0; 3.9題include using namespace std;int main ( ) int a,b,c; coutabc; if(ab) if(bc) coutmax=c; else coutmax=b; else if (ac) coutmax=c; else coutmax=a; coutendl;return 0; 3.9題另一解#inclu

6、de using namespace std;int main ( ) int a,b,c,temp,max ; coutabc; temp=(ab)?a:b; /* 將a和b中的大者存入temp中 */ max=(tempc)?temp:c; /* 將a和b中的大者與c比較，最大者存入max */ coutmax=maxendl; return 0; 3.10題#include using namespace std;int main ( ) int x,y;精品. coutx; if (x1) y=x; coutx=x, y=x=y; else if (x10) / 1x10 y=2*x-

7、1; coutx=x, y=2*x-1=y; else / x10 y=3*x-11; coutx=x, y=3*x-11=y; coutendl;return 0;3.11題#include using namespace std;int main () float score; char grade; coutscore; while (score100|score0) coutscore; switch(int(score/10) case 10: case 9: grade=a;break; case 8: grade=b;break; case 7: grade=c;break; c

8、ase 6: grade=d;break; default:grade=e; coutscore is score, grade is gradeendl; return 0;3.12題#include using namespace std;int main ()精品.long int num; int indiv,ten,hundred,thousand,ten_thousand,place; /*分別代表個(gè)位,十位,百位,千位,萬(wàn)位和位數(shù)*/ coutnum; if (num9999) place=5; else if (num999) place=4; else if (num99)

9、place=3; else if (num9) place=2; else place=1; coutplace=placeendl; /計(jì)算各位數(shù)字 ten_thousand=num/10000; thousand=(int)(num-ten_thousand*10000)/1000; hundred=(int)(num-ten_thousand*10000-thousand*1000)/100; ten=(int)(num-ten_thousand*10000-thousand*1000-hundred*100)/10; indiv=(int)(num-ten_thousand*10000

10、-thousand*1000-hundred*100-ten*10); coutoriginal order:; switch(place) case 5:coutten_thousand,thousand,hundred,ten,indivendl; coutreverse order:; coutindivtenhundredthousandten_thousandendl; break; case 4:coutthousand,hundred,ten,indivendl; coutreverse order:; coutindivtenhundredthousandendl; break

11、; case 3:couthundred,ten,indivendl; coutreverse order:; coutindivtenhundredendl; break; case 2:coutten,indivendl; coutreverse order:; coutindivtenendl; break; case 1:coutindivendl; coutreverse order:;精品. coutindivendl; break; return 0; 3.13題#include using namespace std;int main () long i; /i為利潤(rùn) floa

12、t bonus,bon1,bon2,bon4,bon6,bon10; bon1=100000*0.1; /利潤(rùn)為10萬(wàn)元時(shí)的獎(jiǎng)金 bon2=bon1+100000*0.075; /利潤(rùn)為20萬(wàn)元時(shí)的獎(jiǎng)金 bon4=bon2+100000*0.05; /利潤(rùn)為40萬(wàn)元時(shí)的獎(jiǎng)金 bon6=bon4+100000*0.03; /利潤(rùn)為60萬(wàn)元時(shí)的獎(jiǎng)金 bon10=bon6+400000*0.015; /利潤(rùn)為100萬(wàn)元時(shí)的獎(jiǎng)金 couti; if (i=100000) bonus=i*0.1; /利潤(rùn)在10萬(wàn)元以內(nèi)按10%提成獎(jiǎng)金 else if (i=200000) bonus=bon1+(i-1

13、00000)*0.075; /利潤(rùn)在10萬(wàn)元至20萬(wàn)時(shí)的獎(jiǎng)金 else if (i=400000) bonus=bon2+(i-200000)*0.05; /利潤(rùn)在20萬(wàn)元至40萬(wàn)時(shí)的獎(jiǎng)金 else if (i=600000) bonus=bon4+(i-400000)*0.03; /利潤(rùn)在40萬(wàn)元至60萬(wàn)時(shí)的獎(jiǎng)金 else if (i=1000000) bonus=bon6+(i-600000)*0.015; /利潤(rùn)在60萬(wàn)元至100萬(wàn)時(shí)的獎(jiǎng)金 else bonus=bon10+(i-1000000)*0.01; /利潤(rùn)在100萬(wàn)元以上時(shí)的獎(jiǎng)金 coutbonus=bonusendl; re

14、turn 0; 3.13題另一解#include using namespace std;int main ()long i; float bonus,bon1,bon2,bon4,bon6,bon10; int c; bon1=100000*0.1; bon2=bon1+100000*0.075; bon4=bon2+200000*0.05; bon6=bon4+200000*0.03;精品. bon10=bon6+400000*0.015; couti; c=i/100000; if (c10) c=10; switch(c) case 0: bonus=i*0.1; break; cas

15、e 1: bonus=bon1+(i-100000)*0.075; break; case 2: case 3: bonus=bon2+(i-200000)*0.05;break; case 4: case 5: bonus=bon4+(i-400000)*0.03;break; case 6: case 7: case 8: case 9: bonus=bon6+(i-600000)*0.015; break; case 10: bonus=bon10+(i-1000000)*0.01; coutbonus=bonusendl; return 0;3.14題#include using na

16、mespace std;int main ()int t,a,b,c,d; coutabcd; couta=a, b=b, c=c,d=db) t=a;a=b;b=t; if (ac) t=a; a=c; c=t; if (ad) t=a; a=d; d=t; if (bc) t=b; b=c; c=t; if (bd) t=b; b=d; d=t; if (cd) t=c; c=d; d=t; coutthe sorted sequence:endl; couta, b, c, dendl; return 0;精品. 3.15題#include using namespace std;int

17、 main ()int p,r,n,m,temp; coutnm; if (nm) temp=n; n=m; m=temp; /把大數(shù)放在n中, 小數(shù)放在m中 p=n*m; /先將n和m的乘積保存在p中, 以便求最小公倍數(shù)時(shí)用 while (m!=0) /求n和m的最大公約數(shù) r=n%m; n=m; m=r; couthcf=nendl; coutlcd=p/nendl; / p是原來(lái)兩個(gè)整數(shù)的乘積 return 0; 3.16題#include using namespace std;int main ()char c; int letters=0,space=0,digit=0,other

18、=0; coutenter one line:=a & c=a & c=0 & c=9) digit+; else other+; coutletter:letters, space:space, digit:digit, other:otherendl;精品. return 0; 3.17題#include using namespace std;int main ()int a,n,i=1,sn=0,tn=0; coutan; while (i=n) tn=tn+a; /賦值后的tn為i個(gè)a組成數(shù)的值 sn=sn+tn; /賦值后的sn為多項(xiàng)式前i項(xiàng)之和 a=a*10; +i; couta

19、+aa+aaa+.=snendl; return 0; 3.18題#include using namespace std;int main ()float s=0,t=1; int n; for (n=1;n=20;n+) t=t*n; / 求n! s=s+t; / 將各項(xiàng)累加 cout1!+2!+.+20!=sendl; return 0; 3.19題#include using namespace std;int main ()int i,j,k,n; coutnarcissus numbers are:endl; for (n=100;n1000;n+) i=n/100; j=n/10

20、-i*10; k=n%10;精品. if (n = i*i*i + j*j*j + k*k*k) coutn ; coutendl;return 0; 3.20題#include using namespace std; int main() const int m=1000; / 定義尋找范圍 int k1,k2,k3,k4,k5,k6,k7,k8,k9,k10; int i,a,n,s; for (a=2;a=m;a+) / a是21000之間的整數(shù)，檢查它是否為完數(shù) n=0; / n用來(lái)累計(jì)a的因子的個(gè)數(shù) s=a; / s用來(lái)存放尚未求出的因子之和，開(kāi)始時(shí)等于a for (i=1;ia;

21、i+) / 檢查i是否為a的因子 if (a%i=0) / 如果i是a的因子 n+; / n加1，表示新找到一個(gè)因子 s=s-i; / s減去已找到的因子，s的新值是尚未求出的因子之和 switch(n) / 將找到的因子賦給k1,.,k10 case 1: k1=i; break; / 找出的笫1個(gè)因子賦給k1 case 2: k2=i; break; / 找出的笫2個(gè)因子賦給k2 case 3: k3=i; break; / 找出的笫3個(gè)因子賦給k3 case 4: k4=i; break; / 找出的笫4個(gè)因子賦給k4 case 5: k5=i; break; / 找出的笫5個(gè)因子賦給k

22、5 case 6: k6=i; break; / 找出的笫6個(gè)因子賦給k6 case 7: k7=i; break; / 找出的笫7個(gè)因子賦給k7 case 8: k8=i; break; / 找出的笫8個(gè)因子賦給k8 case 9: k9=i; break; / 找出的笫9個(gè)因子賦給k9 case 10: k10=i; break; / 找出的笫10個(gè)因子賦給k10 精品. if (s=0) / s=0表示全部因子都已找到了 couta is a 完數(shù)endl; cout1) coutk1,1表示a至少有2個(gè)因子 if (n2) cout,2表示至少有3個(gè)因子，故應(yīng)再輸出一個(gè)因子 if (n

23、3) cout,3表示至少有4個(gè)因子，故應(yīng)再輸出一個(gè)因子 if (n4) cout,5) cout,6) cout,7)cout,8)cout,9)cout,k10; coutendlendl; return 0; using namespace std; int main()3.20題另一解#include int m,s,i; for (m=2;m1000;m+) s=0; for (i=1;im;i+) if (m%i)=0) s=s+i; if(s=m) coutm is a完數(shù)endl; coutits factors are:; for (i=1;im;i+) if (m%i=0)

24、 couti ; coutendl; return 0; 3.20題另一解#include using namespace std;int main() int k11; int i,a,n,s; for (a=2;a=1000;a+) n=0;精品. s=a; for (i=1;ia;i+) if (a%i)=0) n+; s=s-i; kn=i; / 將找到的因子賦給k1k10 if (s=0) couta is a 完數(shù)endl; coutits factors are:; for (i=1;in;i+) coutki ; coutknendl; return 0; 3.21題#incl

25、ude using namespace std;int main() int i,t,n=20; double a=2,b=1,s=0; for (i=1;i=n;i+) s=s+a/b; t=a; a=a+b; / 將前一項(xiàng)分子與分母之和作為下一項(xiàng)的分子 b=t; / 將前一項(xiàng)的分子作為下一項(xiàng)的分母 coutsum=sendl; return 0; 3.22題#include using namespace std;int main() int day,x1,x2; day=9; x2=1; while(day0) x1=(x2+1)*2; / 第1天的桃子數(shù)是第2天桃子數(shù)加1后的2倍 x2

26、=x1;精品. day-; couttotal=x1endl; return 0; 3.23題#include #include using namespace std;int main() float a,x0,x1; couta; / 輸入a的值 x0=a/2; x1=(x0+a/x0)/2; do x0=x1; x1=(x0+a/x0)/2; while(fabs(x0-x1)=1e-5); coutthe square root of a is x1endl; return 0; 3.24題#include using namespace std;int main() int i,k;

27、 for (i=0;i=3;i+) / 輸出上面4行*號(hào) for (k=0;k=2*i;k+) cout*; / 輸出*號(hào) coutendl; /輸出完一行*號(hào)后換行 for (i=0;i=2;i+) / 輸出下面3行*號(hào) for (k=0;k=4-2*i;k+) cout*; / 輸出*號(hào) coutendl; / 輸出完一行*號(hào)后換行 return 0; 3.25題精品.#include using namespace std;int main() char i,j,k; /* i是a的對(duì)手;j是b的對(duì)手;k是c的對(duì)手*/ for (i=x;i=z;i+) for (j=x;j=z;j+)

28、if (i!=j) for (k=x;k=z;k+) if (i!=k & j!=k) if (i!=x & k!=x & k!=z) couta-i b-j c-kendl; return 0; 4.1題#include using namespace std;int main() int hcf(int,int); int lcd(int,int,int); int u,v,h,l; cinuv; h=hcf(u,v); couth.c.f=hendl; l=lcd(u,v,h); coutl.c.d=lu) t=u;u=v;v=t; while (r=u%v)!=0) u=v; v=r;

29、 return(v); int lcd(int u,int v,int h) return(u*v/h); 精品.4.2題#include #include using namespace std;float x1,x2,disc,p,q;int main()void greater_than_zero(float,float); void equal_to_zero(float,float); void smaller_than_zero(float,float); float a,b,c; coutabc; disc=b*b-4*a*c; coutroot:0) greater_than_

30、zero(a,b); coutx1=x1,x2=x2endl; else if (disc=0) equal_to_zero(a,b); coutx1=x1,x2=x2endl; else smaller_than_zero(a,b); coutx1=p+qiendl; coutx2=p-qi0時(shí)方程的根 */ x1=(-b+sqrt(disc)/(2*a); x2=(-b-sqrt(disc)/(2*a); void equal_to_zero(float a,float b) /* 定義一個(gè)函數(shù)，用來(lái)求disc=0時(shí)方程的根 */ x1=x2=(-b)/(2*a); void smalle

31、r_than_zero(float a,float b) /* 定義一個(gè)函數(shù)，用來(lái)求disc0時(shí)方精品.程的根 */ p=-b/(2*a); q=sqrt(-disc)/(2*a); 4.3題#include using namespace std;int main() int prime(int); /* 函數(shù)原型聲明 */ int n; coutn; if (prime(n) coutn is a prime.endl; else coutn is not a prime.endl; return 0; int prime(int n) int flag=1,i; for (i=2;in/

32、2 & flag=1;i+) if (n%i=0) flag=0; return(flag); 4.4題#include using namespace std;int main() int fac(int); int a,b,c,sum=0; coutabc; sum=sum+fac(a)+fac(b)+fac(c); couta!+b!+c!=sumendl; return 0; int fac(int n)精品. int f=1; for (int i=1;i=n;i+) f=f*i; return f; 4.5題#include #include using namespace std

33、;int main() double e(double); double x,sinh; coutx; sinh=(e(x)+e(-x)/2; coutsinh(x)=sinhendl; return 0; double e(double x) return exp(x); 4.6題#include #include using namespace std;int main()double solut(double ,double ,double ,double ); double a,b,c,d; coutabcd; coutx=solut(a,b,c,d)=1e-5); return(x)

34、;4.7題#include #include using namespace std;int main()void godbaha(int); int n; coutn; godbaha(n); return 0;void godbaha(int n) int prime(int); int a,b; for(a=3;a=n/2;a=a+2) if(prime(a) b=n-a; if (prime(b) coutn=a+bendl; int prime(int m)int i,k=sqrt(m); for(i=2;ik) return 1; else return 0;4.8題#includ

35、e using namespace std;int main() int x,n; float p(int,int); coutnx;精品. coutn=n,x=xendl; coutpn(x)=p(n,x)endl; return 0; float p(int n,int x) if (n=0) return(1); else if (n=1) return(x); else return(2*n-1)*x*p(n-1),x)-(n-1)*p(n-2),x)/n); 4.9題#include using namespace std;int main()void hanoi(int n,cha

36、r one,char two,char three); int m; coutm; coutthe steps of moving m disks:endl; hanoi(m,a,b,c); return 0;void hanoi(int n,char one,char two,char three) /將n個(gè)盤從one座借助two座，移到three座 void move(char x,char y); if(n=1) move(one,three); else hanoi(n-1,one,three,two); move(one,three); hanoi(n-1,two,one,three); void move(char x,char y) coutxyendl;4.10題#include using namespace std;int main()精品.void convert(int n); int number; coutnumber; coutoutput:endl; if (number0) cout-; numb