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

第一章

1.5題

#include<iostream>

usingnamespacestd;

intmain()

(

coutvv”This“vv”is”;

cout?,,a',?,,C-H-H-

cout?Hprogram.H;

return0;

1.6題

#include<iostream>

usingnamespacestd;

intmain()

(

inta,b,c;

a=10;

b=23;

c=a+b;

cout?na+b=n;

cout?c;

cout?endl;

return0;

}

1.7七題

#include<iostream>

usingnamespacestd;

intmain()

(

inta,b,c;

intf(intx,inty,intz);

cin?a?b?c;

c=fifa,b,c);

cout?c?endl;

return0;

}

intf(intx,inty,intz)

(

intm;

if(x<y)m=x;

elsem=y;

if(z<m)m=z;

retum(m);

}

1.8題

#include<iostream>

usingnamespacestd;

intmain()

(

inta,b,c;

cin?a?b;

c=a+b;

cout?,,a4-b="?a+b?endl;

return0;

}

1.9題

#include<iostream>

usingnamespacestd;

intmain()

{

inta,b,c;

intadd(intx,inty);

cin?a?b;

c=add(a,b);

cout?na+b=n?c?endl;

return0;

)

intadd(intx,inty)

{intz;

z=x+y;

return(z);

}

2.3題

#include<iostream>

usingnamespacestd;

intmain()

{charcl='a',c2='b',c3='c',c4='\10r,c5='\l16;

cout?c1<<c2?c3?,\n,;

cout?,,\t\bn?c4?,\t'?c5?'\n,;

return0;

}

2.4題

#include<iostream>

usingnamespacestd;

intmain()

{charcl='C,c2='+',c3='+'；

cout?nIsay:\HH?cl?c2?c3?,\n,;

cout?n\t\tH?"Hesays:\"C++isveryinteresting!\""?

return0;

}

2.7題

#include<iostream>

usingnamespacestd;

intmain()

{inti,j,m,n;

i=8;

j=10;

m=++i+j++;

n=(++i)+("Hj)+m;

cout?i?,\t,?j?'\t,?m?,\t,?n?endl;

return0;

)

2.8題

#include<iostream>

usingnamespacestd;

intmain()

{charcl-C,c2-h\c3-i\c4='n\c5-a*;

c1+=4;

c2+=4;

c3+=4;

c4+=4;

c5+=4;

cout?npasswordis:n?c1?c2?c3?c4?c5?end1;

return0;

}

3.2題

#include<iostream>

#include<iomanip>

usingnamespacestd;

intmain()

{floath,r,l,s,sq,vq,vz;

constfloatpi=3.1415926;

cout?npleaseenterr,h:n;

cin?r?h;

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;

cout?setiosflags(ios::fixed)?setiosflags(ios::right)

?setprecision(2);

cout?nl=H?setw(lO)?l?endl;

cout?Ms=H?setw(10)?s?endl;

cout?nsq="?setw(lO)?sq?endl;

cout?nvq=n?setw(10)?vq?endl;

cout?nvz=n?setw(10)?vz?endl;

return0;

}

3.3題

#include<iostream>

usingnamespacestd;

intmain()

{floatc,f;

coutw”請(qǐng)輸入一個(gè)華氏溫度:“；

cin?f;

c=(5.0/9.0)*(A32);〃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:"wcwendl;

return0;

}；

3.4題

#include<iostream>

usingnamespacestd;

intmain()

{charcl,c2;

coutw”請(qǐng)輸入兩個(gè)字符cl,c2:M;

cl=getchar();〃將輸入的第一個(gè)字符賦給cl

c2=getchar();〃將輸入的第二個(gè)字符賦給c2

coutvv”用putchar函數(shù)輸出結(jié)果為:";

putchar(c1);

putchar(c2);

cout?endl;

cout?"用cout語(yǔ)句輸出結(jié)果為:";

cout?c1?c2?endl;

return0;

}

3.4題另一?解

#include<iostream>

usingnamespacestd;

intmain()

{charcl,c2;

cout?”請(qǐng)輸入兩個(gè)字符cl,c2:M;

cl=getchar();〃將輸入的第一個(gè)字符賦給cl

c2=getchar();〃將輸入的第二個(gè)字符賦給c2

coutvv”用putchar函數(shù)輸出結(jié)果為:“；

putchar(cl);

putchar(44);

putchar(c2);

cout?endl;

coutvv”用cout語(yǔ)句輸出結(jié)果為:“；

cout?c1?H,H?c2?endl;

return0;

}

3.5題

#include<iostream>

usingnamespacestd;

intmain()

{charcl,c2;

intil,i2;〃定義為整型

coutvv”請(qǐng)輸入兩個(gè)整數(shù)il,i2:n;

cin?il?i2;

cl=il;

c2=i2;

coutw”按字符輸出結(jié)果為:“vvclw"，”v〈c2vvendl;

return0;

}

3.8題

#include<iostream>

usingnamespacestd;

intmain()

{inta=3,b=4,c=5,x,y;

cout?(a+b>c&&b==c)?endl;

cout?(a||b+c&&b-c)?endl;

cout?(!(a>b)&&!c||l)?endl;

cout?(!(x=a)&&(y=b)&&0)?endl;

cout?(!(a+b)+c-l&&b+c/2)?endl;

return0;

}

3.9題

include<iostream>

usingnamespacestd;

intmain()

{inta,b,c;

cout?npleaseenterthreeintegernumbers:"；

cin?a?b?c;

if(a<b)

ifi[b<c)

cout?Hmax=H?c;

else

cout?nmax=H?b;

elseif(a<c)

cout?Hmax=H?c;

else

cout?nmax=M?a;

cout?endl;

return0;

}

3.9題另一解

#include<iostream>

usingnamespacestd;

intmain()

{inta,b,c,temp,max;

cout?npleaseenterthreeintegernumbers:"；

cin?a?b?c;

temp=(a>b)?a:b;/*將a和b中的大者存入temp中*/

max=(temp>c)?temp:c;/*將a和b中的大者與c比較，最大者存入max

*/

cout?nmax=,,?max?endl;

return0;

|

3.10題

#include<iostream>

usingnamespacestd;

intmain()

{intx,y;

cout?nenterx:n;

cin?x;

if(x<l)

{y=x;

cout?Hx="?x?n,y=x=H?y;

}

elseif(x<10)//l<x<10

{y=2*x-l;

cout〈v”x="vvxvv”，y=2*x-1=H?y;

)

else//x>10

{y=3*x-ll;

cout?nx="?x?n,y=3*x-ll-*?y;

cout?endl;

return0;

}

3.11題

#include<iostream>

usingnamespacestd;

intmain()

{floatscore;

chargrade;

cout?npleaseenterscoreofstudent:";

cin?score;

while(score>100||score<0)

{cout?Hdataerror,enterdataagain.";

cin?score;

}

switch(int(score/l0))

{case10:

case9:grade=,A,;break;

case8:grade-B';break;

case7:grade-C*;break;

case6:grade='D';break;

default:grade-E';

}

cout?nscoreisn?score?n,gradeis"?grade?endl;

return0;

}

3.12題

#include<iostream>

usingnamespacestd;

intmain()

{longintnum;

intindiv,ten,hundred,thousand,tenthousand,place;

/*分別代表個(gè)位，十位，百位，千位，萬(wàn)位和位數(shù)

*/

cout?nenteraninteger(0-99999):u;

cin?num;

if(num>9999)

place=5;

elseif(num>999)

place=4;

elseif(num>99)

place=3;

elseif(num>9)

place=2;

elseplace=l;

cout?nplace=,,?place?endl;

〃計(jì)算各位數(shù)字

ten_thousand=num/10000;

thousand=(int)(num-ten_thousand*10000)/1000;

hundred=(int)(num-ten_thousand*l0000-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*10);

cout?noriginalorder:

switch(place)

{case

5:cout?ten_thousand?H,"?thousand?,,；,?hundred?,,,"?ten?"；,?indiv?en

dl；

cout?"reverseorder:n;

cout?indiv?ten?hundred?thousand?ten_thousand?endl;

break;

case4:cout?thousand?,,,''<<hundred?,,,,,?ten?*7，?indiv?endl;

cout?"reverseorder:";

cout?indiv?ten?hundred?thousand?endl;

break;

case3:cout?hundred?,,,',<<ten?",,,<<indiv?endl;

cout?"reverseorder:n;

cout?indiv?ten?hundred?endl;

break;

case2:cout?ten?H,"?indiv?endl;

cout?"reverseorder:";

cout?indiv?ten?endl;

break;

caseI:cout?indiv?endl;

cout?"reverseorder:H;

cout?indiv?endl;

break;

}

return0;

}

313題

#include<iostream>

usingnamespacestd;

intmain()

{longi;//i為利潤(rùn)

floatbonus,bon1,bon2,bon4,bon6,bon10;

bonl=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

bonl0=bon6+400000*0.015;〃利潤(rùn)為100萬(wàn)元時(shí)的獎(jiǎng)金

cout?"enteri:H;

cin?i;

if(i<=100000)

bonus=i*0.1;〃利潤(rùn)在10萬(wàn)元以?xún)?nèi)按10%提成獎(jiǎng)金

elseif(i<=200000)

bonus=bon1+(i-l00000)*0.075;〃利潤(rùn)在10萬(wàn)元至20萬(wàn)時(shí)的獎(jiǎng)金

elseif(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

elseif(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

elseif(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

cout?nbonus=M?bonus?endl;

return0;

}

3.13題另解

#include<iostream>

usingnamespacestd;

intmain()

{longi;

floatbonus,bon1,bon2,bon4,bon6,bon10;

intc;

bonl=100000*0.1;

bon2=bon1+100000*0.075;

bon4=bon2+200000*0.05;

bon6=bon4+200000*0.03;

bon10=bon6+400000*0.015;

cout?nenteri:";

cin?i;

c=i/100000;

if(c>10)c=10;

switch(c)

{case0:bonus=i*0.1;break;

case1:bonus=bonl+(i-100000)*0.075;break;

case2:

case3:bonus=bon2+(i-200000)*0.05;break;

case4:

case5:bonus=bon4+(i-400000)*0.03;break;

case6:

case7:

case8:

case9:bonus=bon6+(i-600000)*0.015;break;

case10:bonus=bon10+(i-1000000)*0.01;

)

cout?nbonus="?bonus?endl;

return0;

)

3.14題

#include<iostream>

usingnamespacestd;

intmain()

(intt,a,b,c,d;

cout?nenterfournumbers:H;

cin?a?b?c?d;

cout?na=n?a?n,b=H?b?H,c=,,?c?,,,d=,,?d?endl;

if(a>b)

{t=a;a=b;b=t;}

if(a>c)

{t=a;a=c;c=t;}

if(a>d)

{t=a;a=d;d=t;}

if(b>c)

{t=b;b=c;c=t;}

if(b>d)

{t=b;b=d;d=t;}

if(c>d)

{t=c;c=d;d=t;}

cout?nthesortedsequence:H?endl;

cout?a?H,H?b?M,H?c?n,n?d?endl;

return0;

)

3.15題

#include<iostream>

usingnamespacestd;

intmain()

{intp,r,n,m,temp;

cout?npleaseentertwopositiveintegernumbersn,m:,r

cin?n?m;

if(n<m)

{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;

cout?uHCF="?n?endl;

cout<<uLCD=n?p/n?endl;//p是原來(lái)兩個(gè)整數(shù)的乘積

return0;

3.16題

#include<iostream>

usingnamespacestd;

intmain()

{charc;

intletters=0,space=0,digit=0,other=0;

cout?nenteroneline::H?endl;

while((c=getchar())!=,\n,)

{if(c>=fa'&&cv=N||c>=A&&c<='Z,)

letters-H-;

elseif(c=**)

space++;

elseif(c>=O&&c<=9)

digit++;

else

other++；

)

cout?,,letter:n?letters<<",space:n?space?n,digit:n?digit?H,

other:"?other?endl;

return0;

}

3.17題

#include<iostream>

usingnamespacestd;

intmain()

{inta,n,i=1,sn=0,tn=0;

cout?na,n=:H;

cin?a?n;

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；

cout?na+aa+aaa+...=,,?sn?endl;

return0;

)

3.18題

#include<iostream>

usingnamespacestd;

intmain()

{floats=0,t=l;

intn;

fbr(n=l;n<=20;n++)

(

t=t*n;〃求n!

s=s+t;//將各項(xiàng)累加

)

cout?H1!+2!+...+20!-,?s?endl;

return0;

)

3.19題

#include<iostream>

usingnamespacestd;

intmain()

{intij,k,n;

cout?Hnarcissusnumbersare:n?endl;

for(n=100;n<1000;n++)

{i=n/100;

j=n/10-i*10;

k=n%10;

if(n=i*i*i+j*j*j+k*k*k)

cout?n?MM;

)

cout?endl;

return0;

}

3.20題

#include<iostream>

usingnamespacestd;

intmain()

(constintm=1000;//定義尋找范圍

intkI,k2,k3,k4,k5,k6,k7,k8,k9,k10;

inti,a,n,s;

fbr(a=2;a<=m;a++)〃a是2-1000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=l;i<a;i++)//檢查i是否為a的因子

if(a%i=0)//如果i是a的因子

{n++;〃11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kl,..”kl0

{case1:

kl=i;break;//找出的笫1個(gè)因子賦給kl

case2:

k2=i;break;//找出的笫2個(gè)因子賦給k2

case3:

k3=i;break;//找出的笫3個(gè)因子賦給k3

case4:

k4=i;break;//找出的笫4個(gè)因子賦給k4

case5:

k5=i;break;//找出的笫5個(gè)因子賦給k5

case6:

k6=i;break;//找出的笫6個(gè)因子賦給k6

case7:

k7=i;break;//找出的笫7個(gè)因子賦給k7

case8:

k8=i;break;//找出的笫8個(gè)因子賦給k8

case9:

k9=i;break;//找出的笫9個(gè)因子賦給k9

case10:

kl0=i;break;//找出的笫10個(gè)因子賦給klO

}

if(s==O)//s=0表示全部因子都已找到了

{cout?a?"isa完數(shù)”vvendl;

cout?'*itsfactorsare:“；

if(n>l)cout?kl?",n?k2;//n>l表示a至少有2個(gè)因子

if(n>2)cout?H,"?k3;//n>2表示至少有3個(gè)因子，故應(yīng)再輸出一個(gè)因子

if(n>3)cout<v","vvk4;〃n>3表示至少有4個(gè)因子，故應(yīng)再輸出?個(gè)因子

if(n>4)coutvv”,”vvk5;//以下類(lèi)似

if(n>5)cout?\"?k6;

if(n>6)cout?n,"?k7;

if(n>7)cout?H,n?k8;

if(n>8)cout?,,,,,?k9;

if(n>9)cout?M,M?klO;

cout?endl?endl;

return0;

}

3.20題另一解

#include<iostream>

usingnamespacestd;

intmain()

{intm,s,i；

for(m=2;m<1000;m-H4-)

{s=0;

for(i=l;i<m;i++)

if((m%i)==0)s=s+i;

if(s==m)

{cout?m?"isa完數(shù)”vvendl;

cout?nitsfactorsare:"；

for(i=l;i<m;i++)

if(m%i==0)cout?i?nM;

cout?endl;

}

)

return0;

}

3.20題另一解

#include<iostream>

usingnamespacestd;

intmain()

{intk[ll]；

inti,a,n,s;

for(a=2;a<=1000;a-H-)

{n=0;

s=a;

for(i=l;i<a;i++)

if((a%i)==0)

{n++；

s=s-i;

k[n]=i;//將找到的因子賦給k[l]—k[10]

)

if(s=0)

{cout?a?Misa完數(shù)”vvendl;

cout?"itsfactorsare:";

for(i=l;i<n;i++)

cout?k[i]?nH;

cout?k[n]?endl;

return0;

3.21題

#include<iostream>

usingnamespacestd;

intmain()

{inti,t,n=20;

doublea=2,b=l,s=0;

for(i=];iv=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

}

cout?nsum=H?s?endl;

return0;

)

3.22題

#include<iostream>

usingnamespacestd;

intmain()

{intday,xl,x2;

day=9;

x2=l;

while(day>0)

{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=xl;

day—;

}

cout?ntotal=,,?x1?endl;

return0;

)

3.23題

#include<iostream>

#include<cmath>

usingnamespacestd;

intmain。

{floata,x0,xl;

cout?nenterapositivenumber:";

cin?a;//輸入a的值

x0=a/2;

xl=(x0+a/x0)/2;

do

{xO=xl;

xl=(x0+a/x0)/2;

}

while(fabs(xO-xl)>=le-5);

cout?nThesquarerootofn?a?His"?xl?endl;

return0;

)

3.24題

#include<iostream>

usingnamespacestd;

intmain()

{inti,k;

for(i=0;i<=3;i-H-)//輸出上面4行*號(hào)

{for(k=0;k<=2*i;k++)

cout?M*M;//輸出*號(hào)

cout?endl;〃輸出完一行*號(hào)后換行

)

for(i=0;i<=2;i-H-)//輸出下面3行*號(hào)

{for(k=0;k<=4-2*i;k++)

cout?M*H;//輸出*號(hào)

cout?endl;//輸出完一行*號(hào)后換行

}

return0;

}

3.25題

#includev沁stream〉

usingnamespacestd;

intmain()

{chari,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++)

for0=X;jv='Z;j")

if(i!=j)

for(k=,X,;k<='Z';k++)

if(i!=k&&j!=k)

if(i!=X&&k!='X'&&k!='Z')

cout?nA-n?i?HB-n<<j?HC-M?k?endl;

return0;

4.1題

#include<iostream>

usingnamespacestd;

intmain()

{inthcf{int,int);

intlcd(int,int,int);

intu,v,h,l;

cin?u?v;

h=hcf{u,v);

cout?nH.C.F="?h?endl;

l=lcd(u,v,h);

cout?,,L.C.D=n?K<endl;

return0;

)

inthcfifintu,intv)

{intt,r;

if(v>u)

{t=u;u=v;v=t;}

while((r=u%v)!=O)

{u=v;

v=r;}

retum(v);

}

intlcd(intu,intv,inth)

{retum(u*v/h);

4.2題

#include<iostream>

#include<math.h>

usingnamespacestd;

floatxl,x2,disc,p,q;

intmain()

{voidgreater_than_zero(float,float);

voidequal_to_zero(float,float);

voidsmaller_than_zero(float,float);

floata,b,c;

cout?ninputa,b,c:";

cin?a?b?c;

disc=b*b-4*a*c;

cout?nroot:H?endl;

if(disc>0)

greater_than_zero(a,b);

cout?nx1=n?x1?,,,x2=,,?x2?endl;

}

elseif(disc==O)

{equal_to_zero(a,b);

cout?nxl=n?xK<n,x2=n?x2?endl;

}

else

{smaller_than_zero(a,b);

cout?,,xl=,'?p?n+,,?q?"i,,?endl;

cout?,,x2-,?p?,,-,,?q?"i"?endl;

}

return0;

)

voidgreater_than_zero(floata,floatb)/*定義-個(gè)函數(shù)，用來(lái)求disc>0時(shí)方

程的根*/

{xl=(-b+sqrt(disc))/(2*a);

x2=(-b-sqrt(disc))/(2*a);

}

voidequal_to_zero(floata,floatb)/*定義,個(gè)函數(shù)，用來(lái)求disc=0時(shí)方程

的根*/

{

x1=x2=(-b)/(2*a);

}

voidsmaller_than_zero(floata,floatb)/*定義個(gè)函數(shù)，用來(lái)求disc〈0時(shí)方

程的根*/

{

p=-b/(2*a);

q=sqrt(-disc)/(2*a);

}

4.3題

#include<iostream>

usingnamespacestd;

intmain()

(intprime(int);/*函數(shù)原型聲明*/

intn;

cout?ninputaninteger:M;

cin?n;

if(prime(n))

cout?n?Hisaprime."?endl;

else

cout?n?Misnotaprime.u?endl;

return0;

|

intprime(intn)

{intflag=l,i;

for(i=2;i<n/2&&flag==l;i++)

if(n%i=0)

flag=0;

retum(flag);

)

4.4題

#include<iostream>

usingnamespacestd;

intmain()

{intfac(int);

inta,b,c,sum=0;

cout?nentera,b,c:M;

cin?a?b?c;

sum=sum+fac(a)+fac(b)+fac(c);

cout?a?M!+"?b?,,!+"?c?H!=,,?sum?endl;

return0;

}

intfac(intn)

{intf^l;

for(inti=l;i<=n;i++)

f=f*i;

returnf;

)

4.5題

#include<iostream>

#include<cmath>

usingnamespacestd;

intmain()

{doublee(double);

doublex,sinh;

cout?nenterx:H;

cin?x;

sinh=(e(x)+e(-x))/2;

cout?nsinh("?x?M)=H?sinh?endl;

return0;

}

doublee(doublex)

{returnexp(x);}

4.6題〃牛頓迭代法

#include<iostream>

#include<cmath>

usingnamespacestd;

intmain()

{doublesolut(double,double,double,double);

doublea,b,c,d;

cout?ninputa,b,c,d:n;

cin?a?b?c?d;

cout?,,x-,?solut(a,b,c,d)?endl;

return0;

}

doublesolut(doublea,doubleb,doublec,doubled)

{doublex=l,x0,f,fl;

do

{x0=x;

f=((a*xO+b)*xO+c)*xO+d;

fl=(3*a*x0+2*b)*x0+c;

x=x0-f7fl;

}

while(fabs(x-xO)>=le-5);

retum(x);

|

intGcd_2(inta,intb)//歐兒里德算法求a,b的最大公約數(shù){if(a<=0||b<=0)//預(yù)防

錯(cuò)誤return0;inttemp;while(b>0)//b總是表示較小的那個(gè)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(gè)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}returna;}

4.7題

#include<iostream>

#include<cmath>

usingnamespacestd;

intmain()

{voidgodbaha(int);

intn;

cout?ninputn:";

cin?n;

godbaha(n);

return0;

)

voidgodbaha(intn)

{intprime(int);

inta,b;

fbr(a=3;a<=n/2;a=a+2)

{if(prime(a))

{b=n-a;

if(prime(b))

cout?n?n=M?a?**+n?b?endl;}

intprime(intm)

{inti,k=sqrt(m);

fbr(i=2;i<=k;i4-+)

if(m%i==0)break;

if(i>k)return1;

elsereturn0;

}

4.8題//遞歸法

#include<iostream>

usingnamespacestd;

intmain()

{intx,n;

floatp(int,int);

cout?ninputn&x:n;

cin?n?x;

cout?nn=n?n?H,x=M?x?endl;;

cout?uPH?n?H(x)=H?p(n,x)?endl;

return0;

}

floatp(intn,intx)

{if(n==0)

retum(l);

elseif(n==l)

retum(x);

else

retum(((2*n-1)*x-p((n-1),x)-(n-1)*p((n-2),x))/n);

}

4.9題〃漢諾塔問(wèn)題

#include<iostream>

usingnamespacestd;

intmain()

{voidhanoi(intn,charone,chartwo,charthree);

intm;

cout?ninputthenumberofdiskes:"；

cin?m;

cout?nThestepsofmoving"?m?"disks:n?endl;

hanoi(m/A,,,B,；C,);

return0;

}

voidhanoi(intn,charone,chartwo,charthree)

〃將n個(gè)盤(pán)從one座借助two座，移到three座

{voidmove(charx,chary);

if{n==l)move(one,three);

else

{hanoi(n-1,one,three,two);

move(one,three);

hanoi(n-1,two,one,three);

}

}

voidmove(charx,chary)

{cout?x?n—>n?y?endl;}

4.10題

#include<iostream>

usingnamespacestd;

intmain()

{voidconvert(intn);

intnumber;

cout?ninputaninteger:";

cin?number;

cout?noutput:H?endl;

if(number<0)

{cout?H-H;

number=-number;

)

convert(number);

cout?endl;

return0;

)

voidconvert(intn)〃感覺(jué)根本想不出的么

{inti;

charc;

if((i=n/10)!=0)

convert(i);

c=n%10+'0';

cout?nH?c;

)

4.11題

#include<iostream>

usingnamespacestd;

intmain()

{intf(int);

intn,s;

cout?ninputthenumbern:";

cin?n;

s=f(n);

cout?nTheresultis"?s?endl;

return0;

}

intfifintn)

(；

if(n=l)

return1;

else

return(n*n+f(n-l));

4.12題

#include<iostream>

#include<cmath>

usingnamespacestd;

#defineS(a,b,c)(a+b+c)/2

#defineAREA(a,b,c)sqrt(S(a,b,c)*(S(a,b,c)-a)*(S(a,b,c)-b)*(S(a,b,c)-c))

intmain()

{floata,b,c;

cout?ninputa,b,c:";

cin?a?b?c;

if(a+b>c&&a+c>b&&b+c>a)

cout?narea=M?AREA(a,b,c)?endl;

else

cout?nItisnotatriangle!n?endl;

return0;

)

4.14題

#include<iostream>

usingnamespacestd;

//#defineLETTER1

intmain()

{charc;

cin?c;

#ifLETTER

iRc>=n&&c<=,z,)

c=c-32;

#else

iRc>=A&&c<=Z)

c=c+32;

#endif

cout?c?endl;

return0;

)

4.15題

#include<iostream>

usingnamespacestd;

#defineCHANGE1

intmain()

{charch[40];

cout?ninputtext:**?endl;;

gets(ch);

#if(CHANGE)

{for(inti=0;i<40;i++)

{if(ch[i]!士\0’)

if(ch[i]>=,a'&&ch[i]<'z,||ch[i]>'A,&&ch[i]<,Z,)

ch[i]+=l;

elseif(ch[i]=,z*||ch[i]=,Z,)

ch[i]-=25;

}

}

#endif

cout?noutput:n?endl?ch?endl;

return0;

}

4.16題file

#include<iostream>

usingnamespacestd;

inta;

intmain()

{externintpower(int);

intb=3,c,d,m;

cout?nenteranintegeraanditspowerm:u?endl;

cin?a?m;

c=a*b;

cout?a?H*n?b?H="?c?endl;

d=power(m);

cout?a?n**,,?m?*-"?d?endl;

return0;

}

416題file

externinta;

intpower(intn)

{inti,y=l;

fbr(i=l;i<=n;iH-4-)

y*=a;

returny;

5.1題

#include<iostream>

usingnamespacestd;

intmain()

{cout〈v2vv-；

for(inti=3;i<=100;i++)

{boolt=true;

fbr(inta=2;a<i;a++)if(i%a=0){t=false;break;}

if(t)cout?i?f

return0;

}

#include<iostream>

#include<iomanip>

usingnamespacestd;

#include<math.h>

intmain()

{inti,j,n,a[101];

for(i=l;i<=100;i++)

a[i]=i；

a[l]=O;

for(i=2;i<sqrt(100);i-?-+)

for(j=i+l；j<=100;j++)

{if(a[i]!=O&&a[j]!=O)

if(a[j]%a[i]=O)

aU]=O；}

cout?endl;

for(i=1,n=O;i<=1()0;i++)

{if(a[i]!=O)

{cout<<setw(5)?a[i]?M**;

n++;}

if(n=10)

{cout?endl;

n=0;}

}

cout?endl;

return0;

)

5.2題

#include<iostream>

usingnamespacestd;

//#include<math.h>

intmain()

{inti,j,min,temp,a[l1];

cout?nenterdata:,,?endl;

fbr(i=l;i<=10;i++)

{cout?,,a[M?i?,,]=M;

cin?a[i];〃輸入10個(gè)數(shù)

)

cout?endl?nTheoriginalnumbers:"?endl;;

fbr(i=1;i<=10;i-H-)

cout?a[i]?n//輸出這10個(gè)數(shù)

cout?endl;;

fbr(i=l;i<=9;i++)//以下8行是對(duì)?10個(gè)數(shù)排序

{min=i;

for(j=i+1;j<=10;j++)

if(a[min]>a[j])min=j;

temp=a[i];〃以下3行將a[i+l]?a[10]中最小者與a[i]對(duì)換

a[i]=a[min];

a[min]=temp;

}

cout?endl?nThesortednumbers:"?endl;

for(i=l;i<=10;i++)〃輸出已排好序的10個(gè)數(shù)

cout?a[i]?"M;

cout?endl;

return0;

}

#include<iostream>

usingnamespacestd;

intmain()

{inta[10],e,k,t;

for(inti=0;i<10;i++){cin?a[i];}

fbr(inti=0;i<9;i++){k=i;for(t=i+l;t<10;t+4-){if(a[i]<a[t])k=t;}e=a[i]；a[i]=a[k];a[k]=e;}

for(inti=0;i<10;i++){cout?a[i]?,

return0;

)

#include<iostream>

#include<iomanip>

usingnamespacestd;

intmain()

{inta[10][10];

fbr(inti=0;i<10;i++){fbr(intt=0;t<10;t++)a[i][t]=0;}

fbr(inti=0;i<10;i++)a[i][i]=l;

for(inti=0;i<10;i++)a[i][0]=l;

fbr(inti=2;i<10;i++){fbr(intt=1;t<=i-1;t++)a[i][t]=a[i-1][t-1]+a[i-1][t];}

fbr(inti=0;i<10;i++){for(intt=0;t<10;t++){if(a[i][t]!=0)cout?a[i][t]?,*;}cout?endl;}

return0;

}

我的第一個(gè)帶指針的C++程序，呵呵??

#include<iostream>

usingnamespacestd;

intmain()

{char*p;inti=0;

charc[100];

gets(c);

P=C；

while(*p!='\0'){i++;p++;}

cout?i;

return0;

}

5.3題

#include<iostream>

usingnamespacestd;

intmain()

{inta[3][3],sum=0;

intij;

cout?nenterdata:"?endl;;

for(i=0;i<3;i++)

for(j=0;j<3;j++)

cin?a[i][j];

for(i=0;i<3;i++)

sum=sum+a[i][i];

cout?Hsum=M?sum?endl;

return0;

}

5.4題

#include<iostream>

usingnamespacestd;

intmain()

{inta[l1]={1,4,6,9,13,16,19,28,40,100};

intnum,ij;

cout?narraya:n?endl;

for(i=0;i<10;i++)

cout?a[i]?"**;

cout?endl;;

cout?ninsertdata:"；

cin?num;

if(num>a[9])

a[10]=num;

else

{for(i=0;i<10;i++)

{if(a[i]>num)

{for(j=9;j>=i;j-)

a[j+l]=a[j];

a[i]=num;

break;

cout?nNow,arraya:n?endl;

for(i=O;i<ll;i-H-)

cout?a[i]?n

cout?endl;

return0;

}

5.5題

#include<iostream>

usingnamespacestd;

intmain()

{constintn=5;

inta[n],i,temp;

cout?nenterarraya:"?endl;

fbr(i=0;i<n;i++)

cin?a[i];

cout?narraya:H?endl;

fbr(i=0;i<n;i4-+)

cout?a[i]?nn;

for(i=0;i<n/2;i++)〃循環(huán)的作用是將對(duì)稱(chēng)的元素的值互換

{temp=a[i];

a[i]=a[n-i-l];

a[n-i-l]=temp;

)

cout?endl?MNow,arraya:n?endl;

fbr(i=0;i<n;i4-+)

cout?a[i]?n

cout?endl;

return0;

)

5.6題

#include<iostream>

#include<iomanip>

usingnamespacestd;

intmain()

{constintn=ll;

inti,j,a[n][n];

fbr(i=l;i<n;i++)

{a[i][i]=l；

a[i][l]=l；

}

fbr(i=3;i<n;i++)

for(j=2;j<=i-l;j++)

a[i]U]=a[i-l]U-l]+a[i-l]U]；

for(i=l;i<n;i-H-)

{fbr(j=l;j<=i;j-H-)

cout?setw(5)?a[i][j]?M

cout?endl;

)

cout?endl;

return0;

}

5.7題

#include<iostream>

usingnamespacestd;

intmain()

{constintn=4,m=5;〃假設(shè)數(shù)組為4行5列

inti,j,a[n][m],max,maxj;

boolflag;

fbr(i=0;i<n;i++)〃輸入數(shù)組

for(j=O;j〈m;j++)

cin?a[i]|j];

fbr(i=0;i<n;i++)

{max=a[i][0];maxj=0;

for(j=0;j<m;j4-+)〃找出第i行中的最大數(shù)

if(a[i][j]>max)

{max=a[i][j];〃將本行的最大數(shù)存放在max中