《c++程序設(shè)計》譚浩強(qiáng)課后習(xí)題答案.doc

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1、第一章1.5題#include using namespace std;int main() coutThisis; coutaC+; coutprogram.; return 0;1.6題#include using namespace std;int main() int a,b,c; a=10; b=23; c=a+b; couta+b=; coutc; coutendl; return 0;1.7七題#include using namespace std;int main() int a,b,c; int f(int x,int y,int z); cinabc; c=f(a,b,c

2、); coutcendl; return 0;int f(int x,int y,int z) int m; if (xy) m=x; else m=y; if (zm) m=z; return(m); 1.8題#include using namespace std;int main() int a,b,c; cinab; c=a+b; couta+b=a+bendl; return 0;1.9題#include using namespace std;int main() int a,b,c; int add(int x,int y); cinab; c=add(a,b); couta+b

3、=cendl; return 0;int add(int x,int y)int z; z=x+y; return(z);2.3題#include using namespace std;int main() char c1=a,c2=b,c3=c,c4=101,c5=116; coutc1c2c3n; couttbc4tc5n; return 0;2.4題#include using namespace std;int main() char c1=C,c2=+,c3=+; coutI say: c1c2c3; coutttHe says: C+ is very interesting! n

4、; return 0;2.7題#include using namespace std;int main()int i,j,m,n; i=8; j=10; m=+i+j+; n=(+i)+(+j)+m; coutitjtmtnendl; return 0;2.8題#include using namespace std;int main()char c1=C, c2=h, c3=i, c4=n, c5=a; c1+=4; c2+=4; c3+=4; c4+=4; c5+=4; coutpassword is:c1c2c3c4c5endl; return 0;3.2題#include #incl

5、ude 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=setw(10)sqendl; coutvq

6、=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請輸入兩個字符c1,c2:; c1=getchar(); /將輸入的第一個字符賦給c1 c2=getchar(); /

7、將輸入的第二個字符賦給c2 cout用putchar函數(shù)輸出結(jié)果為:; putchar(c1); putchar(c2); coutendl; cout用cout語句輸出結(jié)果為:; coutc1c2endl; return 0;3.4題另一解#include using namespace std;int main ( )char c1,c2; cout請輸入兩個字符c1,c2:; c1=getchar(); /將輸入的第一個字符賦給c1 c2=getchar(); /將輸入的第二個字符賦給c2 cout用putchar函數(shù)輸出結(jié)果為:; putchar(c1); putchar(44); p

8、utchar(c2); coutendl; cout用cout語句輸出結(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)endl; cout(a|b+c & b-c

9、)endl; 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#include using namespace std;in

10、t 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; coutx; if (x1) y=x; coutx=x, y=x=y; else if (x10) / 1x10 y=2*x-1; coutx=x, y=2*x-1=y; else

11、/ 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; case 6: grade=D;break; defaul

12、t:grade=E; coutscore is score, grade is gradeendl; return 0;3.12題#include using namespace std;int main ()long int num; int indiv,ten,hundred,thousand,ten_thousand,place; /*分別代表個位,十位,百位,千位,萬位和位數(shù) */ coutnum; if (num9999) place=5; else if (num999) place=4; else if (num99) place=3; else if (num9) place=

13、2; else place=1; coutplace=placeendl; /計算各位數(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-thousand*1000-hundred*100-ten

14、*10); coutoriginal order:; switch(place) case 5:coutten_thousand,thousand,hundred,ten,indiven dl; coutreverse order:; coutindivtenhundredthousandten_thousandendl; break; case 4:coutthousand,hundred,ten,indivendl; coutreverse order:; coutindivtenhundredthousandendl; break; case 3:couthundred,ten,indi

15、vendl; coutreverse order:; coutindivtenhundredendl; break; case 2:coutten,indivendl; coutreverse order:; coutindivtenendl; break; case 1:coutindivendl; coutreverse order:; coutindivendl; break; return 0; 3.13題#include using namespace std;int main () long i; /i為利潤 float bonus,bon1,bon2,bon4,bon6,bon1

16、0; bon1=100000*0.1; /利潤為10萬元時的獎金 bon2=bon1+100000*0.075; /利潤為20萬元時的獎金 bon4=bon2+100000*0.05; /利潤為40萬元時的獎金 bon6=bon4+100000*0.03; /利潤為60萬元時的獎金 bon10=bon6+400000*0.015; /利潤為100萬元時的獎金 couti; if (i=100000) bonus=i*0.1; /利潤在10萬元以內(nèi)按10%提成獎金 else if (i=200000) bonus=bon1+(i-100000)*0.075; /利潤在10萬元至20萬時的獎金 e

17、lse if (i=400000) bonus=bon2+(i-200000)*0.05; /利潤在20萬元至40萬時的獎金 else if (i=600000) bonus=bon4+(i-400000)*0.03; /利潤在40萬元至60萬時的獎金 else if (i=1000000) bonus=bon6+(i-600000)*0.015; /利潤在60萬元至100萬時的獎金 else bonus=bon10+(i-1000000)*0.01; /利潤在100萬元以上時的獎金 coutbonus=bonusendl; return 0; 3.13題另一解#include using n

18、amespace 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; bon10=bon6+400000*0.015; couti; c=i/100000; if (c10) c=10; switch(c) case 0: bonus=i*0.1; break; case 1: bonus=bon1+(i-100000)*0.075; b

19、reak; 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 namespace std;int main ()int t,a,b,c,

20、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; 3.15題#include using namespace std;int main ()int p,r,n,m,temp; coutnm; if (

21、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是原來兩個整數(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=0; coutenter one line:=a & c=A & c=0

22、& c=9) digit+; else other+; coutletter:letters, space:space, digit:digit, other:otherendl; 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a組成數(shù)的值 sn=sn+tn; /賦值后的sn為多項式前i項之和 a=a*10; +i; couta+aa+aaa+.=snendl; return 0; 3.18題#include

23、 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-i*10; k=n%10; if (n = i*i*i + j*j*j + k*

24、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用來累計a的因子的個數(shù) s=a; / s用來存放尚未求出的因子之和，開始時等于a for (i=1;ia;i+) / 檢查i是否為a的因子 if (a%i=0) / 如果i是a的因子 n+; /

25、 n加1，表示新找到一個因子 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個因子賦給k1 case 2: k2=i; break; / 找出的笫2個因子賦給k2 case 3: k3=i; break; / 找出的笫3個因子賦給k3 case 4: k4=i; break; / 找出的笫4個因子賦給k4 case 5: k5=i; break; / 找出的笫5個因子賦給k5 case 6: k6=i; break; / 找出的笫6個因子賦給k6 case

26、7: k7=i; break; / 找出的笫7個因子賦給k7 case 8: k8=i; break; / 找出的笫8個因子賦給k8 case 9: k9=i; break; / 找出的笫9個因子賦給k9 case 10: k10=i; break; / 找出的笫10個因子賦給k10 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個因子 if (n2) cout,2表示至少有3個因子，故應(yīng)再輸出一個因子 if (n3) cout,3表示至少有4個因子，故應(yīng)再輸出一個因子 if (n4) cout,5) co

27、ut,6) cout,7)cout,8)cout,9)cout,k10; coutendlendl; return 0; 3.20題另一解#include using namespace std; int main() 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) couti ; coutendl; return 0; 3.20題另一解#include

28、using namespace std;int main() int k11; int i,a,n,s; for (a=2;a=1000;a+) n=0; 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#include using namespace std;int main() int i,t,n=20;

29、double a=2,b=1,s=0; for (i=1;i=n;i+) s=s+a/b; t=a; a=a+b; / 將前一項分子與分母之和作為下一項的分子 b=t; / 將前一項的分子作為下一項的分母 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=x1; day-; couttotal=x1endl; return 0; 3.23題#incl

30、ude #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; for (i=0;i=3;i+) / 輸出上面4行*號 for (k=0;k=2*i;k+) cout

31、*; / 輸出*號 coutendl; /輸出完一行*號后換行 for (i=0;i=2;i+) / 輸出下面3行*號 for (k=0;k=4-2*i;k+) cout*; / 輸出*號 coutendl; / 輸出完一行*號后換行 return 0; 3.25題#include using namespace std;int main() char i,j,k; /* i是a的對手;j是b的對手;k是c的對手*/ for (i=X;i=Z;i+) for (j=X;j=Z;j+) if (i!=j) for (k=X;k=Z;k+) if (i!=k & j!=k) if (i!=X &

32、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; return(v); int lcd(int u,int v,int h) return(u*v/h); 4

33、.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_zero(a,b); coutx1=x1,x2=x2endl; else if (disc=0) equal_to_

34、zero(a,b); coutx1=x1,x2=x2endl; else smaller_than_zero(a,b); coutx1=p+qiendl; coutx2=p-qi0時方 程的根 */ x1=(-b+sqrt(disc)/(2*a); x2=(-b-sqrt(disc)/(2*a); void equal_to_zero(float a,float b) /* 定義一個函數(shù)，用來求disc=0時方程 的根 */ x1=x2=(-b)/(2*a); void smaller_than_zero(float a,float b) /* 定義一個函數(shù)，用來求disc0時方 程的根 */

35、 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/2 & flag=1;i+) if (n%i=0) flag=0; return(flag); 4.4題#inclu

36、de 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) int f=1; for (int i=1;i=n;i+) f=f*i; return f; 4.5題#include #include using namespace std;int main() double e(double); double x,sinh; coutx; sinh=(e(x

37、)+e(-x)/2; coutsinh(x)=sinhendl; return 0; double e(double x) return exp(x); 4.6題/牛頓迭代法#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);int Gcd_2(int a, int b)/ 歐幾里德算法求a, b的最大公約數(shù) if (a=0 | b

38、 0) /b總是表示較小的那個數(shù)，若不是則交換a，b的值 temp = a % b; /迭代關(guān)系式 a = b; /a是那個膽小鬼，始終跟在b的后面 b = temp; /b向前沖鋒占領(lǐng)新的位置 return a; 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; i

39、f (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/遞歸法#include using namespace std;int main() int x,n; float p(int,int); coutnx; 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 r

40、eturn(2*n-1)*x-p(n-1),x)-(n-1)*p(n-2),x)/n); 4.9題/漢諾塔問題#include using namespace std;int main()void hanoi(int n,char 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個盤從one座借助two座，移到three座 void mov

41、e(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()void convert(int n); int number; coutnumber; coutoutput:endl; if (number0) cout-; number=-number; convert(number); coutendl; return 0; void convert(int n) /感覺根本想不出的么