實(shí)驗(yàn)教程編程答案-副本

1、實(shí)驗(yàn)二 輸入/輸出與順序結(jié)構(gòu)4. 編程題 輸入學(xué)生的語文、數(shù)學(xué)、英語、物理4門課程的成績，計(jì)算該學(xué)生的總成績和平均成績并輸出。#include <iostream.h>void main() double eng, chin,math,phy,sum,aver;cout<<"please input 4 scores: "cin>>eng>>chin>>math>>phy; /輸入成績sum=eng+chin+math+phy; /計(jì)算總成績aver=sum/4; /計(jì)算平均分cout<<&

2、quot;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: "cin>>c1;c2=c1+32;cout<<"c1="<<c1<<&qu

3、ot;tc2="<<c2<<endl;實(shí)驗(yàn)三 選擇結(jié)構(gòu)程序設(shè)計(jì)3. 編程題XX 由鍵盤輸入三個(gè)字符，輸出其中的最大者。XX【源程序】#include <iostream.h>void main() char x,max; cout<<"Please input three characterss: " cin>>x; max=x; cin>>x; if (x>max) max=x; cin>>x; if (x>max) max=x; cout<<"

4、max = "<<max<<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 <iostream.h>void main()int x,y;cout<<"Please input x= "cin>>x;if (x>0)y=1;else if (x=0)y=0;elsey=-1;cout<<"y="<<y<<endl; 計(jì)算獎(jiǎng)金。設(shè)企業(yè)利潤為L，當(dāng)企業(yè)利潤L不超過5000元

5、時(shí)，獎(jiǎng)金為利潤的1.5%，當(dāng)5000L10000元時(shí)，超過5000元部分獎(jiǎng)金為2%（5000元以下仍按1.5%）；當(dāng)10000L20000元，除10000以下的按上述方法計(jì)算外，超過10000元部分按2.5%計(jì)算獎(jiǎng)金；如果20000L50000元，超過20000元部分按3%計(jì)算獎(jiǎng)金；當(dāng)50000L100000元時(shí)，超過50000元部分按3.5%計(jì)算獎(jiǎng)金；當(dāng)L超過100000元時(shí)，超過100000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。【源程序】#include<iostream.h>void main()double L,S;cout<<&qu

6、ot;please input L="cin>>L;if(L<5000) S=L*0.015;else if(L<10000)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<100000)S=175+250+900+(L-50000)*0.035;elseS=175+250+900+1750+(L-100000)*0.04;cout<<"S="

7、<<S<<endl; 輸入年齡，輸出所處人群：9歲以下為兒童，輸出A；1019為少年，輸出B；2029為青年，輸出C；3049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;switch(age/10)case 0:cout<<"A-兒童n"break;case 1:cout<<"B-少年n"break

8、;case 2:cout<<"C-青年n"break;case 3:case 4:cout<<"D-中年n"break;default:cout<<"E-老年n"break;實(shí)驗(yàn)四 循環(huán)結(jié)構(gòu)程序設(shè)計(jì)4編程題 輸入n，求1+2+3+n的和。 #include<iostream.h>void main() int i,n;double sum=0;cin>>n;for(i=1;i<=n;i+) sum+=i;cout<<"1+2+3+.+"&

9、lt;<n<<"="<<sum<<endl;實(shí)驗(yàn)五 典型結(jié)構(gòu)程序設(shè)計(jì)2編程（1）輸入10個(gè)字符，輸出其中的最大者。#include<iostream.h>void main()char ch,maxchar;cout<<"please input ten character:"cin>>ch;maxchar=ch;for(int i=1;i<10;i+) cin>>ch; if(ch>maxchar)maxchar=ch;cout<<&quo

10、t;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高度的一半，再落下，再反彈。求它在第10次落地時(shí)，共經(jīng)過多少米？第10次反彈多高？分析：共經(jīng)過： 100*(1+1/2+1/4+1/8-+1/1024) 米第10次：100/1024米#include<iostream.h>void main()double s=1,t=1,sum,t10;int i;for(i=1;i<=10;i+)t=2*t;s=s+1/t;t10=100/t;sum=100*s;cout<<&quo

11、t;sum="<<sum<<"tt10="<<t10<<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)if(x>0)a+=1;if(x<0)b+=1;if(x%2)c+=1;else d+=1;cin>>x;cout<&

12、lt;"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;實(shí)驗(yàn)六 * 輸入10個(gè)學(xué)生的成績，求其平均值，輸出最高成績，并統(tǒng)計(jì)低于平均值的人數(shù)。 * 注： LT是小于的意思，less than */#include <iostream.h>const dou

13、ble LOWER = - 10000;void main()double score10;double highestScore = LOWER;double average = 0;double numLTaverage = 0;/輸入、求最高分、求總分for (int i = 0; i < 10 ; + i)cout << "Please input the score ( " << i + 1 << "/10):"cin >> scorei;if ( highestScore < sco

14、rei ) highestScore = scorei;average += scorei;average /= 10;/求成績低于平均分的人數(shù)for ( i = 0; i < 10; + i)if ( scorei < average ) + numLTaverage ;cout << "The average score is " << average << endl;cout << "The highest score is " << highestScore <<

15、 endl;cout << "The number of LT average is " << numLTaverage << endl;/* * 分別用冒泡法和選擇法對(duì)輸入的10個(gè)整數(shù)按由大到小排序。 * 冒泡法參見課本第四章課后作業(yè)第四題 */#include <iostream.h>void main()int myArray10;int outer, inner;int imax;for ( int i = 0; i < 10 ; + i )cout << "Input a number:&

16、quot; ;cin >> myArrayi;for ( outer = 0 ; outer < 10 ; + outer)/在下標(biāo)位outer到9之間的元素中尋找最大值imax = outer ;for ( inner = outer + 1; inner < 10 ; + inner)if ( myArrayimax < myArrayinner ) imax = inner;/將最大值與下標(biāo)為outer的元素交換int temp = myArrayouter;myArrayouter = myArrayimax;myArrayimax = temp;/輸出f

17、or ( i = 0; i < 10 ; + i )cout << myArrayi << ' 'cout << endl;/* * 使用折半查找法，在給定的數(shù)組中查找某個(gè)數(shù)據(jù)。 */#include <iostream.h>const int N = 10;void main()int myArrayN = 1, 5, 8, 13, 16, 34, 67, 78, 90, 100;int iSearch;cout << "Please tell me the number which you want

18、 to search: " ;cin >> iSearch;int low = 0;int high = N - 1;int mid;/開始二分查找while ( low <= high )mid = ( low + high ) / 2;if ( myArraymid = iSearch ) break;else if ( myArraymid < iSearch )low = mid + 1;elsehigh = mid - 1;/判斷是否找到if ( myArraymid = iSearch )cout << "We have fo

19、und the number." << endl;elsecout << "We havn't found the number." << endl;/* * 按楊輝三角形的規(guī)律打印以下的數(shù)據(jù)（要求只打印出10行）。 * 1 * 1 1 * 1 2 1 * 1 3 3 1 * 1 4 6 4 1 * 1 5 10 10 5 1 * */#include <iostream.h>void main()int yangHui1010;yangHui00 = yangHui10 = yangHui11 = 1;fo

20、r ( int i = 2; i < 10 ; + i )yangHuii0 = yangHuiii = 1;for ( int j = 1 ; j < i ; + j )yangHuiij = yangHuii - 1j + yangHuii - 1j - 1;for ( i = 0 ; i < 10 ; + i )for (int j = 0; j <= i ; + j )cout << yangHuiij << 't'cout << endl;/*XXX 編寫程序統(tǒng)計(jì)某班英語、語文、數(shù)學(xué)3門課程的成績， * 學(xué)生

21、人數(shù)與成績由鍵盤輸入，要求統(tǒng)計(jì)出每門課程全班 * 的總成績和平均成績以及每個(gè)學(xué)生三門課程的總成績和 * 平均成績。 */#include <iostream.h>void main()/由用戶輸入學(xué)生人數(shù)int numStu;cout << "Please input the number of students: "cin >> numStu;/根據(jù)用戶輸入的人數(shù)建立數(shù)組，其中0-2列為三門課成績，最后一列是總分。/最后添加一行用來存儲(chǔ)總分。int (*p)4;p = new intnumStu + 14;/將總分置0pnumStu0

22、= pnumStu1 = pnumStu2 = 0;for ( int i = 0 ; i < numStu ; + i )/將每個(gè)人的總分置0pi3 = 0;cout << "ID is " << i + 1 << ":n"/輸入英語成績cout << "Please input the score of Eng: " ;cin >> pi0;pi3 += pi0;pnumStu0 += pi0;/輸入物理成績cout << "Please in

23、put the score of Phy: " ;cin >> pi1;pi3 += pi1;pnumStu1 += pi1;/輸入英語成績cout << "Please input the score of Mat: " ;cin >> pi2;pi3 += pi2;pnumStu2 += pi2;/輸出for ( i = 0 ; i < numStu ; + i )cout << "ID " << i + 1 << ":n"cout <&

24、lt; "The sum is " << pi3 ;cout << ".tThe average is " << pi3/3.0 ;cout << endl;cout << "Eng: sum=" << pnumStu0 << ",average=" << pnumStu0/double(numStu);cout << "nPhy: sum=" << pnumStu1 <

25、< ",average=" << pnumStu1/double(numStu);cout << "nMat: sum=" << pnumStu2 << ",average=" << pnumStu2/double(numStu);delete p;/* * 編寫程序求對(duì)矩陣進(jìn)行轉(zhuǎn)置，即將元素的行列位置交換。 */#include <iostream.h>void main()int myMatrix44;/輸入for ( int i = 0; i <

26、 4; + i )for ( int j = 0; j < 4; + j)cout << '(' << i + 1 << ',' << j + 1 << "):"cin >> myMatrixij;/輸出轉(zhuǎn)置前的數(shù)組cout << "Before exchange:n"for ( i = 0; i < 4; + i )for ( int j = 0; j < 4; + j)cout << myMatrixij

27、<< 't'cout << endl;/轉(zhuǎn)置for ( i = 0 ; i < 4 ; + i)for ( int j = 0 ; j < i ; + j)/* * 編寫程序求兩個(gè)矩陣的乘積，若矩陣ANM與BMK相乘， *則得到矩陣C，其行列數(shù)為N×K。注意A的列數(shù)與B的行數(shù)相同 *，才可以進(jìn)行乘法操作。 */#include <iostream.h>const int N = 3;const int M = 4;const int K = 5;void main()/定義數(shù)組int aNM,bMK,cNK;/輸入數(shù)組A

28、cout << "Matrix A:n"for ( int i = 0; i < N; + i)for ( int j = 0; j < M; + j)cout << '(' << i + 1 << ',' << j + 1 << ")/"cout << '(' << N << ',' << M << "):"cin >&g

29、t; aij;/輸入數(shù)組Bcout << "Matrix B:n"for ( i = 0; i < M; + i)for ( int j = 0; j < K; + j)cout << '(' << i + 1 << ',' << j + 1 << ")/"cout << '(' << M << ',' << K << "):"c

30、in >> bij;/計(jì)算C，并輸出cout << "Matrix C=AXB:n"for ( i = 0; i < N; + i)for ( int j = 0; j < K; + j)cij = 0;for (int k = 0 ; k < M ; + k)cij += aik * bkj;cout << cij << 't'cout << endl;int temp = myMatrixij;myMatrixij = myMatrixji;myMatrixji = temp;

31、/輸出轉(zhuǎn)置后的數(shù)組cout << "After exchange:n"for ( i = 0; i < 4; + i )for ( int j = 0; j < 4; + j)cout << myMatrixij << 't'cout << endl;實(shí)驗(yàn)七 數(shù)組與指針(一)4編寫程序并上機(jī)調(diào)試運(yùn)行 編寫程序，輸入5個(gè)字符串，輸出其中最大者。要求使用二維字符數(shù)組及字符串函數(shù)。#include <iostream.h>#include<string.h>const int N=

32、3;void main()char aN20;int i,max=0;for(i=0;i<N;i+)cin.getline(ai,20);for(i=1;i<N;i+)if(strcmp(ai,amax)>0)strcpy(amax,ai);cout<<amax<<endl; 編寫程序，將一個(gè)字符串中的數(shù)字字符都刪除。#include <iostream.h>void main()char a20;int i=0,j=0;cout<<”Please input the characters: ”;cin.getline(a,20

33、);for(i=0;ai!='0'i+)if(ai<'0'|ai>'9')aj+=ai;aj='0' cout<<a<<endl;XXXXX 編寫程序，輸入一行字符，統(tǒng)計(jì)其中有多少個(gè)單詞，單詞之間用一個(gè)或多個(gè)空格分隔。#include <iostream.h>void main()char a100;int i,num=0; cout<<"Please input the characters:n"cin.getline(a,100);for(i=0;

34、ai!='0'i+)while(ai=' ')i+;num+;while(ai!=' ')i+;cout<<"The number of the words are: "<<num;實(shí)驗(yàn)八 數(shù)組與指針(二)4編寫程序并上機(jī)調(diào)試運(yùn)行 編寫程序，當(dāng)輸入17（表示星期幾）時(shí)，顯示相應(yīng)的星期的英文名稱，輸入其它整數(shù)時(shí)則顯示錯(cuò)誤信息。#include"iostream.h"const int M = 2;const int N = 3;void main()char *p7 = "M

35、onday", "Tuesday", "Wednesday", "Thirsday","Friday", "Saturday", "Sunday"char num;for(;)cout<<"Please input a number:"cin>>num;if(num < '1' | num >'7')cout<<"Error!"break;cout

36、<<pnum - 49<<endl;實(shí)驗(yàn)九 函數(shù)及其調(diào)用2. 編程題 編寫一個(gè)判斷素?cái)?shù)的函數(shù)，在主函數(shù)中由鍵盤輸入整數(shù)的范圍，并給出在該范圍內(nèi)的所有素?cái)?shù)。源程序?yàn)?int fun(int n);void main() int m1,m2;cout<<"Please input the range of integer numbers:n"cin>>m1>>m2; for(int i=m1;i<m2;i+) if(fun(i) cout<<i<<" t" int fu

37、n(int n) int i; for(i=2;i<n;i+) if(n%i=0) return 0; return n; XXXXXX 編寫一個(gè)函數(shù)，根據(jù)給定的年、月、日輸出該日是該年的第幾天。在主函數(shù)中調(diào)用該函數(shù)并輸出結(jié)果，從鍵盤輸入年、月、日的值。#include<iostream.h>int fun(int,int,int);void main()int year,month,day;cout<<"please input year,month,day:"cin>>year>>month>>day;w

38、hile(1)if(year>0&&month>=1&&month<=12&&day>=1&&day<=31)cout<<year<<"-"<<month<<"-"<<day<<" is the "<<fun(year,month,day)<<"th day of the year!"<<endl;break;els

39、ecout<<"input error!please input again:" cin>>year>>month>>day;int fun(int year,int month,int day)int ds=day;switch(month-1)case 11:ds+=30;case 10:ds+=31;case 9:ds+=30;case 8:ds+=31;case 7:ds+=31;case 6:ds+=30;case 5:ds+=31;case 4:ds+=30;case 3:ds+=31;case 2:if(year

40、%4=0&&year%100!=0|year%400=0) ds+=29;else ds+=28;case 1:ds+=31;return ds; 編寫兩個(gè)函數(shù)分別求2n，n!，在主函數(shù)中調(diào)用這兩個(gè)函數(shù)計(jì)算21×1!+ 22×2!+2n×n!(n<10)，并在主函數(shù)中輸入n的值，輸出結(jié)果。#include<iostream.h>double fun1(int n);double fun2(int n);void main()int n,i;double s=0;cout<<"please input n:&q

41、uot;cin>>n;for(i=1;i<=n;i+)s+=fun1(i)*fun2(i);cout<<"s="<<s<<endl;double fun1(int n)double m=1;for(int i=1;i<=n;i+)m*=2;return m;double fun2(int n)double m=1;for(int i=1;i<=n;i+)m*=i;return m; 編寫函數(shù)求sin(x)，求sin(x)的近似公式為：在主函數(shù)中輸入x的值并調(diào)用該函數(shù)，輸出結(jié)果。#include<iost

42、ream.h>#include<math.h>double fun(double);void main()double x,sum;cout<<"please input x:"cin>>x;sum=fun(x);cout<<"sum= "<<sum<<endl;cout<<"sin"<<x<<"="<<sin(x)<<endl;double fun(double x)doubl

43、e s=0,t=x;for(int i=1;fabs(t)>1e-6;i+=2)s+=t;t=-t*x*x/(i+1)/(i+2);return s;實(shí)驗(yàn)十 函數(shù)與指針2. 編程題 編寫一個(gè)程序，求方程ax2+bx+c=0的根，用三個(gè)函數(shù)分別求出當(dāng)b2-4ac大于0，小于0和等于0時(shí)的根。要求從主函數(shù)輸入a,b,c的值并輸出結(jié)果。源程序?yàn)?方法1：#include<iostream.h>#include<math.h>void root1(double,double,double,double *,double *);double root2(double,dou

44、ble);void root3(double,double,double,double *,double *);void main()double a,b,c,d,x,x1,x2;cout<<"Please input a,b,c="cin>>a>>b>>c;if(a)d=b*b-4*a*c;if(d>0) root1(a,b,d,&x1,&x2); cout<<"x1="<<x1<<"nx2="<<x2<<

45、;endl;else if(d=0) x=root2(a,b);cout<<"x="<<x<<endl;else root3(a,b,d,&x1,&x2); cout<<"x1="<<x1<<"+"<<x2<<"i"<<endl;cout<<"x2="<<x1<<"-"<<x2<<"i

46、"<<endl;else if(b) cout<<"x="<<-c/b<<"n"else if(c)cout<<"方程無根,輸入的a,b,c值有誤！"<<"n"elsecout<<"方程有無窮多解。"<<"n"void root1(double a, double b, double d, double * x1, double * x2)*x1=(-b+sqrt(d)/

47、(2*a);*x2=(-b-sqrt(d)/(2*a);double root2(double a, double b)double x=-b/(2*a);return x;void root3(double a, double b, double d, double * x1, double * x2)*x1=-b/(2*a);*x2=sqrt(-d)/(2*a);方法2：#include<iostream.h>#include <math.h>void root1(double,double,double,double);void root2(double,doub

48、le,double);void root3(double,double,double,double);double x1,x2;void main()double a,b,c,t;cout<<"Please input a,b,c="cin>>a>>b>>c;if(a)t=b*b-4*a*c;if (t>0)root1(a,b,c,t);cout<<"x1="<<x1<<"nx2="<<x2<<"n"e

49、lse if(t=0)root2(a,b,c);cout<<"x1=x2="<<x1<<"n"elseroot3(a,b,c,t);cout<<"x1="<<x1<<"+"<<x2<<"i"<<"n"<<"x2="<<x1<<"-"<<x2<<"i"&

50、lt;<"n"else if(b) cout<<"x="<<-c/b<<"n"else if(c)cout<<"方程無根,輸入的a,b,c值有誤！"<<"n"elsecout<<"方程有無窮多解。"<<"n"void root1(double a,double b,double c,double t)t=sqrt(t);x1=(-b+t)/(2*a);x2=(-b-t

51、)/(2*a);void root2(double a,double b,double c)x1=x2=-b/(2*a);void root3(double a,double b,double c,double t)t=sqrt(-t);x1=-b/(2*a);x2=t/(2*a); 編寫將一個(gè)整數(shù)n轉(zhuǎn)換成字符串的函數(shù)。在主函數(shù)中調(diào)用該函數(shù)并輸出結(jié)果，從鍵盤輸入n的值。例如輸入123，則輸出字符串“123”。n的位數(shù)可以任意。#include<iostream.h>void convert(int n);void main() int number; cout<<&qu

52、ot;please input an integer:" cin>>number; cout<<"the output is:"<<endl; if(number<0) cout<<"-" number=-number; convert(number); cout<<endl; void convert(int n) int i; char c; if(i=n/10)!=0) convert(i); c=n%10+'0' cout<<" &q

53、uot;<<c; 編寫一個(gè)函數(shù)實(shí)現(xiàn)對(duì)數(shù)組長度為n的整型數(shù)組a的升序排序。在主函數(shù)中調(diào)用該函數(shù)并輸出結(jié)果，從鍵盤輸入數(shù)組元素的值。#include<iostream.h>const int N=50;void input(int ,int);void sort(int ,int);void output(int ,int);void main()int aN,n;cout<<"input length of the array:"cin>>n;cout<<"input the arrayn"inp

54、ut(a,n);cout<<"the array before sortedn"output(a,n);sort(a,n);cout<<"the array after sortedn"output(a,n);void input(int x,int n) int i;for(i=0;i<n;i+)cin>>xi;void sort(int x,int n) int i,j,t;for(i=0;i<n-1;i+)for(j=0;j<n-1-i;j+)if(xj>xj+1)t=xj;xj=xj+1;xj+1=t;void output(int x,int n) int i;for(i=0;i<n;i+)cout&l