河北工業(yè)大學(xué)-C++課后習(xí)題實(shí)驗(yàn)習(xí)題

1、精選優(yōu)質(zhì)文檔-傾情為你奉上習(xí)題11.1 書寫一個(gè)簡(jiǎn)單的 C+程序，從而理解 C+程序的結(jié)構(gòu)及特點(diǎn)。該程序只由一個(gè)主函數(shù)組成，主函數(shù)的函數(shù)體中只包含一個(gè)語(yǔ)句：cout<<"This is a C+ program."【1.1 答】#include <iostream.h>void main()cout<<" This is a C+ program." 習(xí)題22.1      寫出 C+標(biāo)識(shí)符的命名規(guī)則

2、，寫出 5個(gè)合法的標(biāo)識(shí)符和 5個(gè)不合法的標(biāo)識(shí)符。2.2      下列符號(hào)哪些是 C+標(biāo)識(shí)符？哪些不是，為什么？2m  a+c  int  i*j  mm  data  n-a  m22  x  -n  a_1【2.2 答】mm  data  m22 

3、; x  a_1 是C+ 標(biāo)識(shí)符，int是是 C+標(biāo)識(shí)符也是關(guān)鍵字2.3      指出下列合法的常量，并說(shuō)明其類型。     65538   1.3   2.1E4 7L  9uL  '9'  'a'  123u【2.3 答】65538 

4、;  整型常量1.3     double型常量2.1E4   double型常量7L     長(zhǎng)整型常量9uL     無(wú)符號(hào)長(zhǎng)整型常量'9'     字符型常量'a'     字符型常量123u    無(wú)符號(hào)整型常量2.4&

5、#160;     寫出符合 C+規(guī)則的 int、 double、char 和字符串4種類型的常量各 2個(gè)。2.5      說(shuō)明字符常量和字符串常量有何區(qū)別？2.6      下列變量的定義是否合法？為什么？ INT  i,j,k;       char c1,c2; 

6、;    int a,b;float f,b; unsigned int m     char: a1,a2,a3;  unsigned float x,y;【2.6 答】和合法。的INT要小寫；少分號(hào)； char后多了冒號(hào)；數(shù)據(jù)類型錯(cuò)誤2.7      將下列數(shù)學(xué)表示式寫成 C+表達(dá)式。（1 ）       

7、;     （2 ）(3)              (4)    【2.7答】（1 ）(2*x*x+3*y*y)/(x-y)（2 ）(-b+sqrt(b*b-4*a*c)/(2*a)    和  (-b-sqrt(b*b-4*a*c)/(2*a)或 (-b+sqrt(

8、b*b-4*a*c)/2/a        和  (-b-sqrt(b*b-4*a*c)/2/a（3 ）1+(1+a/b)/(1-(a/c)（4 ）x/sqrt(fabs(x*x*x+y*y*y+z*z*z)2.8      已有如下變量定義，求下列表達(dá)式的值。double x=1.2,y=8.5;int a=3;x+a%3*(int)(x+y)int a=2,b=3;double x=3.5,y=2.5;(doub

9、le)(a+b)/2+(int)x%(int)yint x=4,y=8;(+x)*(-y)int e=1,f=4,g=2;double m=10.5,n=4.0,k;k=(e+f)/g+sqrt(n)*1.2/g+mdouble x=2.5,y=4.7;int a=7;x+a%3*(int)(x+y)%2/4int a,b;int x;x=(a=2,b=5,a+,b+,a+b)【2.8 答】  1.2 3.5 35 13.7 2.5  9  1.2#include <iostre

10、am.h>void main()    double x=1.2,y=8.5;    int a=3;    cout<<(x+a%3*(int)(x+y)<<endl; 3.5#include <iostream.h>void main()    int a=2,b=3;    double x=3.5,y=2.5;  &#

11、160; cout<<(double)(a+b)/2+(int)x%(int)y)<<endl; 35#include <iostream.h>void main()    int x=4,y=8;    cout<<(+x)*(-y)<<endl;  13.7#include <iostream.h>#include <math.h>void main()   

12、 int e=1,f=4,g=2;    double m=10.5,n=4.0,k;    k=(e+f)/g+sqrt(n)*1.2/g+m;    cout<<k<<endl; 2.5#include <iostream.h>void main()    double x=2.5,y=4.7;    int a=7; 

13、0;  cout<<(x+a%3*(int)(x+y)%2/4)<<endl;  9#include <iostream.h>void main()    int a,b;    int x;    x=(a=2,b=5,a+,b+,a+b);    cout<<x<<endl;習(xí)題33.1 編寫程序求行駛的距離，輸入行駛的速度

14、和時(shí)間，輸出距離。void main()int  s,t,d;cin>>s>>t;d=s*t;cout<<"d="<<d<<endl; 3.2 編寫程序求圓柱體的表面積和體積。程序運(yùn)行時(shí)輸入圓柱的半徑和高，輸出圓柱的表面積和體積。#include<iostream.h>const double PI=3.;void main()     double  r,h,areaf,volume;

15、0;    cin>>r>>h;     areaf=2.*PI*r*h+PI*r*r*2.;     volume=PI*r*r*h;     cout<<"area="<<areaf<<endl;   cout<<"volume="<<volume&l

16、t;<endl;3.3 輸入梯形的上邊長(zhǎng)、下邊長(zhǎng)及高度，計(jì)算梯形的面積。#include<iostream.h>const double PI=3.;void main()double  d1,d2,h,area;cin>>d1>>d2>>h;area=(d1+d2)*h/2.;cout<<"area="<<area<<endl;  3.4 編程求輸入的兩個(gè)字符中的最小者。#include<iostream.h>v

17、oid main()char c1,c2;cin>>c1>>c2;cout<<"c1="<<c1<<"tc2="<<c2<<endl;if(c1>c2)   cout<<"min="<<c2<<endl;else   cout<<"min="<<c1<<endl;3.5 編程求輸入的三個(gè)

18、整數(shù)中的最大值。#include<iostream.h>void main()     int  a,b,c,max;     cin>>a>>b>>c;     max=a;     if(b>max)max=b;     if(c>max)max=c; &

19、#160;   cout<<"max="<<max<<endl;3.6 編寫程序：輸入一個(gè)整數(shù)，若為 0或正數(shù)時(shí)，顯示 :is positive；為負(fù)數(shù)時(shí)，顯示:is negative。#include<iostream.h>void main()     int  a;     cin>>a;   

20、0;  if(a>=0)             cout<<a<<" is positive"<<endl;     else        cout<<a<<" is negative"<<endl

21、;3.7 編程求一元二次方程的實(shí)根：當(dāng)有實(shí)根時(shí)，輸出兩個(gè)實(shí)根；沒(méi)有實(shí)根時(shí)，顯示提示信息。實(shí)驗(yàn)3  p48 例3.83.8 某超市為促進(jìn)銷售，采用了購(gòu)物打折扣的優(yōu)惠方法。設(shè)每位顧客一次購(gòu)物的金額為 b，當(dāng) 1000b2000 時(shí)，給予九五折優(yōu)惠；當(dāng) 2000b3000 時(shí)，給予九折優(yōu)惠；當(dāng) 3000b5000 時(shí)，給予八五折優(yōu)惠；當(dāng) b 5000時(shí)，給予八折優(yōu)惠。編寫程序，輸入顧客的購(gòu)物款，輸出顧客的應(yīng)付款。該題修改：b>=5000, 八折。

22、#include<iostream.h>#include<math.h>void main()      double b;      cin>>b;      switch(int(b)/1000)             case 0:cout<

23、;<"m="<<b<<endl;break;       case 1:cout<<"m="<<b*0.95<<endl;break;       case 2:cout<<"m="<<b*0.90<<endl;break;      &

24、#160;case 3:       case 4:       cout<<"m="<<b*0.85<<endl;break;       case 5:       default :cout<<"m="<<b*0.

25、80<<endl;          3.9 編一程序，將輸入數(shù)據(jù)歸入某一類別，輸出其類別。數(shù)據(jù)分為四類：小于 10、 10 99、 100 999、 1000以上。例如，輸入 358時(shí)，顯示 “358 is 100 to 999”。#include<iostream.h>#include<math.h>void main()    

26、;   int x;       cin>>x;       if(x<10)cout<<x<<" is littler then 10"<<endl;       if(x>=10&&x<=99)cout<<x<<" is

27、 10 to 99"<<endl;       if(x>=100&&x<=999)cout<<x<<" is 100 to 999"<<endl;       if(x=1000)cout<<x<<" is equal 1000"<<endl;    &#

28、160;  if(x>1000)cout<<x<<" is bigger then 1000"<<endl;3.10 編一程序，輸入字符 a時(shí)顯示 America，b 時(shí)顯示Britain， c時(shí)顯示 China， d時(shí)顯示 Denmark，其它字符時(shí)顯示Harland。#include<iostream.h>#include<math.h>void main()    

29、  char c;      cin>>c;      if(c>='A'&&c<='Z')c=c+32;      switch(c)             case 'a':c

30、out<<"America"<<endl;break;       case 'b':cout<<"Britain"<<endl;break;       case 'c':cout<<"China"<<endl;break;      

31、; case 'd':cout<<"Denmark"<<endl;break;       default :cout<<"Harland"<<endl;          3.11 編寫程序，輸入一整數(shù)，當(dāng)其為 0 6范圍內(nèi)的整數(shù)時(shí)，輸出對(duì)應(yīng)的星期： Sunday、Mond

32、ay 、Tuesday、 Wednesday、Thursday 、Friday、 Saturday，為其它整數(shù)時(shí)，輸出error。#include<iostream.h>void main()        int n;        cout<<"please input an integer:"    

33、0;   cin>>n;        switch(n)                    case 0:cout<<"Sunday"break;        

34、;    case 1:cout<<"Monday"break;            case 2:cout<<"Tuesday"break;            case 3:cout<<"Wednesday"brea

35、k;            case 4:cout<<"Thursday"break;            case 5:cout<<"Friday"break;           

36、 case 6:cout<<"Saturday"break;            default:cout<<"error!"<<endl;          實(shí)驗(yàn)一熟悉 Microsft Visual C+ 6.0開(kāi)發(fā)環(huán)境四思考題#include <iostream.h&g

37、t;void main()    int a,b,c;    cout<<"Please input a,b= "     / 輸出字符串    cin>>a>>b;  /輸入兩個(gè)數(shù)據(jù)到變量a、 b中    c=a*b;      /

38、求乘積存入c    cout<<a<<"*"<<b<<=<<c<<endl; / 輸出結(jié)果 實(shí)驗(yàn)二輸入 /輸出與順序結(jié)構(gòu)三 實(shí)驗(yàn)內(nèi)容1. 閱讀程序，寫出運(yùn)行結(jié)果。i+j=15i*j=50a=9     b=5 2. 程序填空   k=i+j   i<<'+'&

39、lt;<j<<'='   a=c   "char"<<c<<"tASCII="<<a 3. 程序改錯(cuò)#include <iostream.h>void main()    double r,s,l;    cout<<"Please input R"   

40、 cin>>r;    s=3.1416*r*r;    l=2.0*3.1416*r;    cout<<"S="<<s<<endl;    cout<<"L="<<l<<endl; 4. 編程題 輸入華氏溫度F，計(jì)算輸出對(duì)應(yīng)的攝氏溫度。由華氏溫度F求攝氏溫度c的公式為：#

41、include <iostream.h>void main()    double f,c;    cout<<"please input F=:"    cin>>f;    c=(f-32)*5/9;   /或c=5.0/9*(f-32);    cout<<"C="<&

42、lt;c<<endl; 輸入學(xué)生的語(yǔ)文、數(shù)學(xué)、英語(yǔ)、物理 4門課程的成績(jī)，計(jì)算該學(xué)生的總成績(jī)和平均成績(jī)并輸出。#include <iostream.h>void main()    double eng, chin,math,phy,sum,aver;    cout<<"please input 4 scores: "    cin>>eng>>chin>>mat

43、h>>phy;      / 輸入成績(jī)    sum=eng+chin+math+phy;      /計(jì)算總成績(jī)    aver=sum/4;                   

44、60; / 計(jì)算平均分    cout<<"Sum="<<sum<<endl<<"tAverage="<<aver<<endl; / 輸出 編寫程序，從鍵盤輸入一個(gè)大寫英文字母，輸出對(duì)應(yīng)的小寫字母。#include <iostream.h>void main()char c1,c2;cout<<"Please input an upper letter: "ci

45、n>>c1;c2=c1+32;cout<<"c1="<<c1<<"tc2="<<c2<<endl; 實(shí)驗(yàn)三選擇結(jié)構(gòu)程序設(shè)計(jì)三 實(shí)驗(yàn)內(nèi)容1.選擇題 C B D C C C D 2.-44  5   992,11 3. 編程題 由鍵盤輸入三個(gè)字符，輸出其中的最大者?！驹闯绦颉?include <iostr

46、eam.h>void main()    char x,max;    cout<<"Please input three characterss: "    cin>>x;    max=x;    cin>>x;       if (x>max) 

47、       max=x;    cin>>x;       if (x>max)        max=x;    cout<<"max = "<<max<<endl; 輸入三角形三邊的長(zhǎng)，求三角形的面積。若輸入的

48、三個(gè)邊能構(gòu)成三角形，則計(jì)算其面積并輸出；否則輸出提示信息?！驹闯绦颉?參見(jiàn)教材 P44例 3.15#include<iostream.h>#include <math.h>void main()    double a,b,c,s,area;    cout<<"Please input a,b,c="    cin>>a>>b>>c;  

49、60; if (a+b<=c|a+c<=b|b+c<=a)    cout<<"cant be a triangle!n"    else            s=(a+b+c)/2;        area=sqrt(s*(s-a)*(s-b)*(s-c)

50、;        cout<<"area="<<area<<endl;      編程求下面符號(hào)函數(shù)值： y = 0   （x=0）  1   （x>0）  -1   （ x<0） 【源程序】#include <io

51、stream.h>void main()    int x,y;    cout<<"Please input x= "    cin>>x;    if (x>0)       y=1;    else if (x=0)    &#

52、160;  y=0;    else       y=-1;    cout<<"y="<<y<<endl; 計(jì)算獎(jiǎng)金。設(shè)企業(yè)利潤(rùn)為 L，當(dāng)企業(yè)利潤(rùn) L不超過(guò) 5000元時(shí)，獎(jiǎng)金為利潤(rùn)的 1.5%，當(dāng) 5000 L10000元時(shí)，超過(guò)5000 元部分獎(jiǎng)金為 2%（ 5000

53、元以下仍按 1.5%）；當(dāng) 10000 L20000元，除10000 以下的按上述方法計(jì)算外，超過(guò) 10000元部分按 2.5%計(jì)算獎(jiǎng)金；如果 20000 L50000元，超過(guò)20000 元部分按3%計(jì)算獎(jiǎng)金；當(dāng) 50000 L元時(shí)，超過(guò)50000 元部分按3.5%計(jì)算獎(jiǎng)金；當(dāng) L超過(guò) 元時(shí)，超過(guò) 元部分按4%計(jì)算獎(jiǎng)金。由鍵盤輸入 L的值，編程計(jì)算相應(yīng)的獎(jiǎng)金并輸出?！驹闯绦颉?include<iostream.h>void

54、main()    double L,S;    cout<<"please input L="    cin>>L;    if(L<5000)       S=L*0.015;    else if(L<10000)    

55、60;  S=75+(L-5000)*0.02;    else if(L<20000)       S=175+(L-10000)*0.025;    else if(L<50000)       S=175+250+(L-20000)*0.03;    else if(L<)  

56、;     S=175+250+900+(L-50000)*0.035;    else       S=175+250+900+1750+(L-)*0.04;    cout<<"S="<<S<<endl;  輸入年齡，輸出所處人群： 9歲以下為兒童，輸出 A； 10 19為少年，

57、輸出 B； 20 29為青年，輸出 C； 30 49為中年，輸出 D； 50以上為老年，輸出 E。【源程序】#include <iostream.h>void main()    int age;    cout<<"Please input age: "    cin>>age;    sw

58、itch(age/10)        case 0:cout<<"A-兒童n"break;    case 1:cout<<"B-少年n"break;    case 2:cout<<"C-青年n"break;    case 3:    case 4:cout

59、<<"D-中年n"break;    default:cout<<"E-老年n"break;      有如下函數(shù)：0 t 11t 22t 33t 4 由鍵盤輸入t 值，計(jì)算S的值?！驹闯绦颉?方法一#include <iostream.h>void main() double t,s;cout<<"please inp

60、ut t= "cin>>t;if(t>0&&t<4)   if(t<1)      s=t*t;   else if(t<2)      s=t*t-1;   else if(t<3)      s=t*t-2*t+1;   els

61、e     s=t*t+4*t-17;   cout<<"s="<<s<<endl;else   cout<<"Error! t cant be less than 0 or more than 4!n" /方法二#include <iostream.h>void main()double t;cout<<"please input t= "ci

62、n>>t;if(t<=0)   cout<<"error, t cant be less than 0!n"else if(t<1)   cout<<"s="<<t*t<<endl;else if(t<2)   cout<<"s="<<t*t-1<<endl;else if(t<3)   cout&

63、lt;<"s="<<t*t-2*t+1<<endl;else if(t<4)   cout<<"s="<<t*t+4*t-17<<endl;else   cout<<"error, t cant be more than 4!n" /方法三#include <iostream.h>void main()double t,s;cout<<"please

64、 input t="cin>>t;if(t<=0)   cout<<"error, t cant be less than 0!n"else if(t<1)      s=t*t;   cout<<"s="<<s<<endl;   else if(t<2)     

65、60;s=t*t-1;   cout<<"s="<<s<<endl;   else if(t<3)      s=t*t-2*t+1;   cout<<"s="<<s<<endl;   else if(t<4)     s=t*t+4*t-17;

66、   cout<<"s="<<s<<endl;  else   cout<<"error, t cant be more than 4!n"習(xí)題33.12 編寫求100 以內(nèi)偶數(shù)之和的程序。#include<iostream.h>void main()        int s=0,i;  

67、0;     for(i=2;i<=100;i=i+2)            s=s+i;        cout<<"s="<<s<<endl;3.13 輸入一實(shí)數(shù) x和一整數(shù) n，求 xn。#include<iostream.h>

68、void main()        double x,m=1;        int n,i;        cout<<"please input an integer and  a real number:"        

69、;cin>>n>>x;        for(i=1;i<=n;i+)            m=m*x;        cout<<"x的n 次冪為:"<<m<<endl;3.14 編程求 1+2+3+

70、.+n之和超過(guò)2000 的第一個(gè)n值及其和。#include<iostream.h>void main()        double s=0;        int n;        for(n=1;n<=100;n+)         

71、           s=s+n;            if(s>2000)break;                cout<<"s="<<s<&l

72、t;"n="<<n<<endl;3.15 求 200以內(nèi)的素?cái)?shù)之和。#include<iostream.h>void main()        int s=0,n,i;        for(n=1;n<200;n=n+2)           

73、         for(i=2;i<n;i+)                if(n%i=0)break;            if(i=n)s=s+n;     &#

74、160;          cout<<"s="<<s<<endl;3.16 計(jì)算 1-3+5-7+-99+101的值。#include<iostream.h>void main()        int s=0,m=1,n;        for(n

75、=1;n<=101;n=n+2)                    s=s+n*m;            m=-m;            

76、0;   cout<<"s="<<s<<endl;3.17 輸入 10個(gè)整數(shù)，求最小值并輸出。#include<iostream.h>void main()        int x,min,i;        cin>>x;       

77、 min=x;        for(i=2;i<=10;i+)                    cin>>x;            if(x<min)min

78、=x;                cout<<”min=”<<min<<endl;3.18 輸入若干學(xué)生成績(jī)，統(tǒng)計(jì)其中 90分以上、 70 89、 60 69、小于 60分的人數(shù)并輸出，當(dāng)輸入的數(shù)小于 0時(shí)結(jié)束。#include<iostream.h>void main()   

79、; double g;    int i,l=0,m=0,n=0,k=0;    cin>>n;    for(i=1;i<=n;i+)           cin>>g;       if(g>=90)l+;   

80、0;   else if(g>=70)m+;       else if(g>=60)n+;       else if(g>=0)k+;       else break;         cout<<”l=”<<l<&

81、lt;”m=”<<m<<”n=”<<n<<”k=”<<k<<endl;3.19 用牛頓迭代法求方程 x3-2x-4=0在0.5 附近的根，當(dāng)前后兩項(xiàng)的差的絕對(duì)值小于 10-6則達(dá)到要求精度。#include<iostream.h>#include<math.h>void main()        double x1,x2=0.5,y1,y2;   &

82、#160;    cout<<" x1  x2  yn"        do                    x1=x2;      

83、0;     y1=x1*x1*x1-2*x1*x1-4;            y2=3*x1*x1-2;            x2=x1-y1/y2;            cout&

84、lt;<"t"<<x1<<"t"<<x2<<"t"<<y1<<endl;        while(fabs(x1-x2)>1e-6);        cout<<"x="<<x1<<endl; 3.20 用迭代法求&#

85、160;，當(dāng)前后兩項(xiàng)的差的絕對(duì)值小于 10-6則達(dá)到要求精度。求平方根的迭代公式為： #include<iostream.h>#include<math.h>void main()        double a,xn0,xn1;        cout<<”please input a positive number:”;     

86、;   cin>>a;        xn0=a/2;        xn1=(xn0+a/xn0)/2;        do                

87、   xn0=xn1;           xn1=(xn0+a/xn0)/2;        while(fabs(xn0-xn1)>=1e-6);        cout<<a<<”平方根=”<<xn1<<endl; 實(shí)驗(yàn)四循環(huán)結(jié)構(gòu)

88、程序設(shè)計(jì)  三 實(shí)驗(yàn)內(nèi)容1選擇題 下面程序的運(yùn)行結(jié)果是_ D _。#include <iostream.h>2閱讀程序，寫出運(yùn)行結(jié)果。下面程序的運(yùn)行結(jié)果是_。 下面程序的運(yùn)行結(jié)果是_a=4_。 下面程序的運(yùn)行結(jié)果是_n=4_。3程序填空 下面程序的功能是依次顯示 100， 80， 60， 40， 20這 5個(gè)數(shù)，請(qǐng)?zhí)羁铡?include <iostream.h>void main()    int i;

89、    for(i=100;_i>=20_;_i-=20_)        cout<<i<<"t"    cout<<endl; 下面程序的功能是計(jì)算xn，請(qǐng)?zhí)羁铡?include <iostream.h>void main()    int n,x;    cout<

90、;<"Please input x,n="    cin>>x>>n;    double y=1;    for(int i=0;i<_n_;i+)        _   y*=x;_;    cout<<y<<endl; 下面程序的功能

91、是計(jì)算1-3+5-7+-99+101的值，請(qǐng)?zhí)羁铡?include <iostream.h>void main()    int i,t,s=0,sign=1;    for(i=1;i<=101;i+=2)            _t=i*sign_;        s+=t;  

92、;      sign=-sign_;        cout<<"s="<<s<<endl;   *  * *  下面程序的功能是輸出以下形式的金字塔圖案：#include <iostream.h>void main()    for(int i=1;i<=4;i

93、+)            for(int j=1;j<=_4-i_;j+)            cout<<' '        for(j=1;j<=_2*i-1_;j+)     &

94、#160;      cout<<'*'        cout<<endl;    4編程題 輸入n ，求1+2+3+n的和。       #include<iostream.h>void main()      &

95、#160;        int i,n;               double sum=0;               cin>>n;     

96、60;         for(i=1;i<=n;i+)                  sum+=i;                  &

97、#160;     cout<<"1+2+3+.+"<<n<<"="<<sum<<endl; 輸入若干個(gè)整數(shù)，求它們的和，遇到 -999時(shí)結(jié)束輸入。       #include<iostream.h>void main()          in

98、t n;          double sum=0;        cin>>n;          for(;n!=-999;)    /while(n!=-999)        &

99、#160;                        sum+=n;                cin>>n;      

100、;                  cout<<"sum="<<sum<<endl;  輸入一整數(shù)，輸出各位數(shù)字之和。如輸入 6228，則輸出 6+2+2+8的和為18 。#include<iostream.h>void main()    double s=

101、0;    int n;    cin>>n;    while(n!=0)                s+=n%10;      n=n/10;        &

102、#160; cout<<"s="<<s<<endl; 輸入一實(shí)數(shù)x 和一整數(shù)n，求 x+x2+x3+xn的值。#include <iostream.h>void main()          int i,j,n;          double x,sum=0,p;  

103、        cin>>n>>x;          for(i=1;i<=n;i+)                         &

104、#160; p=1.;               for(j=1;j<=i;j+)                 p*=x;           &#

105、160;     sum+=p;                      cout<<"x="<<x<<"tn="<<n<<endl;        

106、;  cout<<"sum="<<sum<<endl;   求2 ！+4！ +6!+16！。#include<iostream.h>void main()             int i,j;           &#

107、160; double sum=0,p;             for(i=2;i<=16;i+=2)                             &

108、#160;    p=1;                    for(j=1;j<=i;j+)                   p*=j; 

109、60;                  sum+=p;                        cout<<"2!+4!+6!+.+16!="

110、;<<sum<<endl;   輸入兩個(gè)整數(shù)n 和m，打印 n行星號(hào)，每行 m個(gè)星號(hào)。如果輸入的 n和 m的值為 4 7，則輸出為：* *  *   *#include <iostream.h>void main()          int i,j,n,m;    

111、60;     cin>>n>>m;          for(i=0;i<n;i+)                            

112、;  for(j=0;j<i;j+)                   cout<<" "             for(j=0;j<m;j+)     

113、0;            cout<<"*"            cout<<endl;            求1n+3n+5n+7n+(2m-1)n ，其中m和 n的值

114、從鍵盤輸入。#include <iostream.h>void main()        int i,j,n,m,sum=0,p;        cin>>n>>m;        for(i=1;i<=2*m-1;i+=2)       

115、60;            p=1.;           for(j=1;j<=n;j+)             p*=i;        

116、0;    sum+=p;                cout<<"m="<<m<<"tn="<<n<<endl;        cout<<"sum="<<sum<

117、<endl;循環(huán)結(jié)構(gòu)的繼續(xù) ：1程序填空 以下程序的功能是計(jì)算： s=1+12+123+1234+12345。請(qǐng)?zhí)羁铡?#160;    t=10*t+i      s=s+t 下面程序的功能是輸出符合條件的三位整數(shù)：它是完全平方數(shù)，又有兩位數(shù)字相同，并且統(tǒng)計(jì)個(gè)數(shù)，請(qǐng)?zhí)羁铡?i-n1*100)/10 或(i/10)%10num+j+2編程（1 ）輸入10個(gè)字符，輸出其中的最大者。#include<iostream.h>

118、void main()       char ch,maxchar;       cout<<"please input ten character:"       cin>>ch;       maxchar=ch;     

119、;  for(int i=1;i<10;i+)                               cin>>ch;          

120、0;    if(ch>maxchar)maxchar=ch;                     cout<<"maxchar="<<maxchar<<endl;（2 ）一個(gè)球從100m高度自由落下，每次落地后反彈回原來(lái)高度的一半，再落下，再反彈。求它在第 10

121、次落地時(shí)，共經(jīng)過(guò)多少米？第 10次反彈多高？分析：共經(jīng)過(guò)： 100*(1+1/2+1/4+1/8-+1/1024) 米第10 次：100/1024米#include<iostream.h>void main()    double a,b=100,sum=100;    for(a=1;a<=1;a+)          b=b/2;  &

122、#160;   sum=sum+2*b;        cout<<sum<<'t'<<b; （3）用下列泰勒級(jí)數(shù)求sinx的近似值，x的值從鍵盤輸入，精度要求為10-6。 #include<iostream.h>#include<math.h>void main()       int sign=1,n=1; 

123、      double x=3.0,fenzi=x,fenmu=1.0,equo=1,sum=x;       while(fabs(equo)>=1e-6)                     fenzi*=x*x;  

124、60;           fenmu*=(2*n)*(2*n+1);              sign*=-1;              equo=sign*fenzi/fenmu;  

125、;            sum+=equo;              n+;              cout<<"sin x="<<sum<&

126、lt;endl;（4 ）編寫程序，對(duì)輸入的一批整數(shù)統(tǒng)計(jì)出正數(shù)的個(gè)數(shù)、負(fù)數(shù)的個(gè)數(shù)、奇數(shù)的個(gè)數(shù)、偶數(shù)的個(gè)數(shù)，要求所統(tǒng)計(jì)的整數(shù)由鍵盤輸入，以 0作為輸入數(shù)據(jù)結(jié)束的標(biāo)志。#include<iostream.h>void main()       int a=0,b=0,c=0,d=0,x;       cin>>x;       while(x!=0)

127、                     if(x>0)a+=1;              if(x<0)b+=1;          

128、    if(x%2)c+=1;              else d+=1;              cin>>x;            &

129、#160; cout<<"正數(shù)個(gè)數(shù)="<<a<<endl;       cout<<"負(fù)數(shù)個(gè)數(shù)="<<b<<endl;       cout<<"奇數(shù)個(gè)數(shù)="<<c<<endl;       cout<<"偶數(shù)個(gè)數(shù)="<<d<<endl;（5）用牛頓迭代法求方程 2x3-4x 2+3x6=0在1.5 附近的根，精度要求為 10-6。#include<iostream.h>#include<math.h>void main()