2015年电大数据结构习题复习资料.doc

习题一 （p12）1-4 # include void main( ) cout ”c+语言程序设计”endl; 1-5 # include void main() int i,j,k;cout ijk;cout endli”+”j”+”k”=”i+j+kendl; 习题二 （p47）2-5 / p50 习题2-5 # include # include void main() double a,r; cout endla; r=1.0/2*a*sqrt(10+2*sqrt(5)/5); cout endl正五边形外接圆的半径：rendl;2-6 / p50 习题2-6# include # include void main()double a,b,c,area,s;cout endlabc;s=(a+b+c)/2;area=sqrt(s*(s-a)*(s-b)*(s-c); cout endl三角形的面积：areaendl;2-7 / p50 习题2-7# include # include void main()double x1,y1,z1,x2,y2,z2,d;cout endlx1y1z1;cout endlx2y2z2;d=sqrt(x2-x1)*(x2-x1)+(y2-y1)*(y2-y1)+(z2-z1)*(z2-z1); cout endl两点间的距离：dendl;习题三 （p82）3-3 / p84 习题3-3# include void main()int weekday;cout endlweekday;if (weekday=0)cout sundayendl;else if (weekday=1)cout mondayendl;else if (weekday=2)cout tuesdayendl;else if (weekday=3)cout wednesdayendl;else if (weekday=4)cout thursdayendl;else if (weekday=5)cout fridayendl;else if (weekday=6)cout saturdayendl;else cout 表示星期几的数必须在0-6之间！endl;3-4 / p84 习题3-4# include void main()int a,b,c,d,max;cout abcd;max=a;if (maxb) max=b;if (maxc) max=c;if (maxd) max=d;cout endl;cout 最大的数是：maxendl;3-5 / p84 习题3-5# include void main()int a,b,c,d,temp;cout abcd;cout 排序前：a b c db) temp=a; a=b; b=temp;if (ac) temp=a; a=c; c=temp;if (ad) temp=a; a=d; d=temp;/ 求第二小数存入bif (bc) temp=b; b=c; c=temp;if (bd) temp=b; b=d; d=temp;/ 求第三小的数存入cif (cd) temp=c; c=d; d=temp;cout 排序后：a b c dendl;3-6 / p84 习题3-6# include void main()cout endl |; / 以下3行显示顶部表栏 int i=1;while (i10)cout setw(3)i;i+;cout endl-+-;i=1;while (i10) / 乘数从1 循环到9cout endlsetw(2)i|;int j=1;while (j10) cout setw(3)i*j; / 被乘数从1 循环到9 j+;i+; coutendlendl;3-7 / p84 习题3-7# include void main()int a=1,b=2; double s=0,p;for (int i=1; i=20; i+)p=(double)b/a;s+=p;b+=a; / 产生分子a=b-a; / 产生分母cout 2/1+3/2+5/3+8/5+.=sendl;3-8 / p84 习题3-8# include # define n 15void main()int n; long double s=0.0l,fact=1.0l;for (n=1; n=n; n+) fact*=n; / 计算n! s+=fact;cout 1!+2!+.+n!=sendl;3-9 / p84 习题3-9 / 方法一# include void main()int n1=1,n2=1,n=n1+n2,couter=2; / n1，n2相邻二项的/分母，cout统计项数 double s=2.0,s0;while (s0=1.0/n)=0.000001)s+=s0;n1=n2;n2=n;n=n1+n2;couter+; coutendlendl; cout 累加和：sendl; cout 累加项数：couterendl;/ 方法二# include void main()int n=2,d=1,k=2; / k统计项数 double s=2.0,s0;while (s0=1.0/n)=0.000001)s+=s0;n+=d;d=n-d;k+; coutendlendl; cout 累加和：sendl; cout 累加项数：kendl;3-10 / p84 习题3-10# include void main()for (int i=1; i=6; i+)cout endl;int j;for (j=1; ji; j+) cout ;for (j=1; j=13-i-i; j+) cout *;cout endlendl;3-11 / p81 习题3-11# include void main() cout setw(6)*; / 输出顶* for (int i=0; i4; i+)cout endl;int j;for (j=0; j4-i; j+) / 输出载边空格cout ;cout *; / 输出左*for (j=0; ji+i+1; j+) / 输出中间空格cout ;cout *; / 输出右*cout endl*;cout endlendl;3-12 / p81 习题3-12# include void main() int i;cout endl |; for (i=0; i=9; i+)cout setw(6)i;cout endl-+;for (i=0; i=10; i+) cout -;for (i=0; i=9; i+)cout endlsetw(5)i|;for (int j=0; j=9; j+)cout setw(6)(i*10+j)*(i*10+j);cout endlendl;3-13 / p81 习题3-13# include # include void main() double x;cout x;int i=1;double n=-x*x,d=1.0*2.0,s=1.0,s0;while (fabs(s0=n/d)=0.000001)s+=s0;i+;n*=(-x*x);d*=(i+i)*(i+i-1);cout 累加求值 cos(x)=sendl;cout 标准函数 cos(x)=cos(x)endl;cout endlendl;习题四 （p107）4-3 / p109 习题4-3 # include # define size 10void main()int asize; int temp,i,max,min;cout 输入sizea0;max=min=0; / 存放最大数和最小数的下标for (i=1; iai;if (aiamax) max=i;/ 保存当前最大数的下标if (aiamin) min=i; / 保存当前最小数的下标cout endl交换前：;for (i=0; isize; i+)cout ai ;temp=a0; a0=amax; amax=temp; / 最大数存/首元素temp=asize-1; asize-1=amin; amin=temp; / 最小数存末元素 cout endl交换后：;for (i=0; isize; i+)cout ai ;cout endlendl;4-4 / p109 习题4-4 方法一：# include void main()int a=23,25,13,68,3,44,11,7,55,36;const int size=sizeof(a)/sizeof(a0); / 数组a的元素/个数 int temp,i;cout endl处理前：;for (i=0; isize; i+)cout ai ;for (i=0; i5; i+) / 找前5大数之一int k=i,j; / k保存大数的下标for (j=5; jak) k=j; temp=ai; ai=ak; ak=temp; / 交换cout endl处理后：;for (i=0; isize; i+)cout ai ;cout endlendl;方法二：# include void main()int a=23,25,13,68,3,44,11,7,55,36;const int size=sizeof(a)/sizeof(a0); int temp,i;cout endl处理前：;for (i=0; isize; i+)cout ai ;for (i=0; i5; i+) / 用选择排序法排前5个数为大数int k=i,j; / k保存当前最大数的下标for (j=i+1; jak) k=j; temp=ai; ai=ak; ak=temp; / 交换cout endl处理后：;for (i=0; isize; i+)cout ai ;cout endlendl;4-5 / p109 习题4-5# include # define size 20void main()int asize,i; cout endl请输入size个正整数: ;for ( i=0; iai;while (ai=0);cout endl处理前: ;for (i=0; isize; i+) cout ai ;cout endl;int j=0; / j为奇数存放下标 for (i=0; isize; i+)if ( ai%2 ) / 条件成立，表示ai为奇数int temp=ai; ai=aj; aj=temp;j+; / 下一个奇数存放位置cout endl处理后: ;for (i=0; isize; i+) cout ai ;cout endlendl;4-6 / p109 习题4-6# include void main() int a=23,25,13,68,3,44,11,7,55,36;const int size=sizeof(a)/sizeof(a0); / 数组a的元素个数 int i,number; cout endlnumber;cout endlendl整数序列: ;for (i=0; isize; i+) / 显示数组元素 cout ai ;cout endlendl输入数: number的序号: ; for (i=0; isize & ai!=number; i+) ; / 查找 cout (isize)?i+1:0)endl; / 表达式中条件成立，表示找到cout endl;4-7 / p109 习题4-7# include # define size 5void main() int asizesize;int i,j,sum=0; cout endl请输入size行size列整型数组: endl;for (i=0; isize; i+)for (j=0; jaij;cout endlendlsize行size列整型矩阵: ;for (i=0; isize; i+) / 显示二维数组元素cout endl;for (j=0; jsize; j+) cout setw(4)aij;for (i=0; isize; i+)for (j=0; jsize; j+)if (i=0 | j=0 | i=size-1 | j=size-1)sum+=aij;cout endlendl矩阵四周元素的合计值: sumendlendl;4-8 / p109 习题 4-8# include # include # define size 10void main()double datasizesize;int i,j,size;cout size;if (size10)cout 行列式的阶数要求在 1-10 之间.endl;return;/ 输入行列式cout 请输入size行size列的行列式：endl;for (i=0; isize; i+)for (j=0; jdataij;/ 输出行列式cout endlendlsize行size列的行列式：endl; for (i=0; isize; i+)cout endl;for (j=0; jsize; j+)cout setw(10)dataij; cout endl;/ 调用函数grmdet()计算行列式的值double grmdet(double asize,int size); cout endlsize行size列的行列式的值为：;cout grmdet(data,size)endl;/ 计算行列式的值函数。以下函数定义参阅1982年12月第1版上海教育出版社出版/ 上海计算技术研究所编电子计算机算法手册第295页算法double grmdet(double asize,int size)int i,j,k,n;double p,t,u;double msize;for (i=0; isize; i+) u=0.0;for (j=0; ju) ) u=abs(aij);if (u1.0e-18) return 0.0;mi=u;if (u!=1)for (j=0; jsize; j+)aij/=u;p=1;for (k=0; ksize-1; k+)n=k;t=abs(akk);for (j=k+1; jsize; j+)if (tabs(akj) t=abs(akj);n=j;if (t1.0e-18) return 0.0;if (n!=k) p=-p;for (i=k; isize; i+) t=aik;aik=ain;ain=t;p*=akk;for (i=k+1; isize; i+) t=aik/akk;for (j=k+1; jsize; j+)aij-=akj*t;t=p*asize-1size-1;for (k=0; ksize; k+)t*=mk;return t;4-9 / p109 习题4-9方法一：# include # include # define size 80void main()char ssize;cout 请输入一个字符串: ;cin.getline(s,80);cout endl原字符串: sendl;char strsize; / 用str存放子串if (strstr(s,boy) / 条件成立，表示原字符串s/有子串boy strcpy(str,strstr(s,boy); / 复制子串boy到str中 int len1=strlen(s); / 原字符串s的长度 int len2=strlen(str); / 子串长度 slen1-len2=0; / 在原字符串s中截断子串 strcat(s,child); / 在s字符串中连接新子串 strcat(s,&str3); / 连接原子串boy后的字符串 cout 替换后串: sendl;elsecout 无子串boy.endlendl;方法二：# include # include void main() cout 请输入一字符串：;char s80;cin.getline(s,80); cout ”原串：”sendl;char *p=strstr(s,boy); / 取子串boy位置if (p) / 条件成立，表示存在子串boyp0=0; / 在原串中截断子串boychar c80; strcpy(c,s); / 复制原串中截断子串boy后剩下的串strcat(c,child); / 连接替换的串childstrcat(c,&p3); / 连接原子串boy后的字符串cout ”新串：”c;elsecout 无子串boy.;cout endl;4-10 / p109 习题4-10 # include # include # define size 80void main()char ssize; cout endl请输入一个字符串: ;cin.getline(s,80);cout endl加密前串: sendl;int i;for (i=1; isize ; i+=2) / 加密char temp=si;si=si-1;si-1=temp; cout 加密后串：sendl;for (i=1; isize ; i+=2) / 解密char temp=si;si=si-1;si-1=temp;cout endl解密后串: sendl;cout endlendl;习题五 （p140）5-3 / p142 习题5-3# include int f(int b,int n) / 计算b的n次方递归函数if (n=0) return 1;return b*f(b,n-1);void main() / 依题意解题时主函/数可以/不必书写，这里给出主函数只是方便 /同学们理解程序(以下相同)int x,n;cout xn;cout x的n次方是: ;cout f(x,n)endl;5-4 / p142 习题5-4# include int arraymin(int d,int size) / 求一维数组元素的最/小值的递归函数int j=size-1; / 默认有size个元素的数组d中,/下标为size-1的元素是最小元素if (size=1) return dj;if (djd0) int temp=dj;dj=d0;d0=temp; return arraymin(d,size-1);void main() / 依题意解题时主函数可以不必书写 # define size 10int datasize,i;cout 请输入size个整数存入数组中:endl;for (i=0; idatai;cout endl输入的数据是：endl;for (i=0; isize; i+)cout datai ;cout endl最小数是：arraymin(data,size)endl;5-5 / p142 习题5-5 头文件 exe5_5.hdouble distance(double,double,double,double);double arround(double ,double ,int);# define size 100/ p142 习题5-5 程序文件 exe5_5.cpp# include exe5_5.h# include double distance(double x1,double y1,double x2, double y2) / 计算两点坐标(x1,y1)和(x2,y2)的距离函数return sqrt(x2-x1)*(x2-x1)+(y2-y1)*(y2-y1);double arround(double x,double y,int p) / 计算多边形周长的函数double s=0;for ( int j=0; jp-1; j+)s+=distance(xj,yj,xj+1,yj+1);s+=distance(xp-1,yp-1,x0,y0);return s;/ p142 习题5-5 主程序文件 exemain.cpp# include # include exe5_5.hvoid main()double xsize,ysize; int maxsize; / 多边形的最大边数cout endlmaxsize;cout 请按顺时针或逆时针顺序输入多边形 ;cout maxsize 个顶点的坐标(格式(x,y): endl;int i;for (i=0; ixiyi;cout 多边形的 maxsize 个顶点坐标为：; for (i=0; imaxsize; i+)cout (xi,yi);cout endl;cout多边形的周长: arround(x,y,maxsize)endl;5-6 / p143 习题5-6 头文件 exe5_6.hvoid selsort(int ,int);# define size 100/ p143 习题5-6 程序文件 sort.cpp# include exe5_6.hvoid selsort(int data,int size)for (int m=0; msize; m+)int j=m;for (int i=m+1; isize; i+)if (dataim)int k=datam;datam=dataj;dataj=k;/ p143 习题5-6 主程序文件 exe5_6.cpp# include # include exe5_6.hvoid main()int maxsize;cout endlmaxsize;int datasize;int i;cout 请输入 maxsize 个待排序的数据: endl;for (i=0; idatai;cout endl排序前: ;for (i=0; imaxsize; i+)cout datai ;selsort(data,maxsize);cout endl排序后: ; for (i=0; imaxsize; i+)cout datai ;cout endl;5-7 / p143 习题5-7 # include # include char * f(char * s1=hello,char * s2=world!);char s80;char * f(char * s1,char * s2)strcpy(s,s1);return strcat(s,s2);void main() / 依题意解题时主函数可以不必书写cout f()endl;cout f(hi,)endl;cout f(hi,friend!)endl;5-8 方法一：#include unsigned power(unsigned n,unsigned x=2)unsigned s=1;for (unsigned i=0; ix; i+)s*=n;return s;void main() / 依题意解题时主函数可以不必书写cout power(2),;cout power(2,0),;cout power(2,1),;cout power(2,2),;cout power(2,3),;cout power(2,4),;cout power(3),;cout power(3,3),;cout power(3,4)endl;方法二：# include unsigned power(unsigned n,unsigned x=2)if (x=0) return 1;bool flag=n=2;for ( unsigned j=2; j1) unsigned s=n;for (unsigned i=1; ix; i+)s*=n;return s;else if (x=1) return n;return 1;5-9 / p143 习题5-9# include template t s(t d,int size)int i;t sum=0;for (i=0; isize; i+)sum+=di*di;return sum;void main() / 依题意解题时主函数可以不必书写int iarray=1,2,3,4,5;double darray=1.1,2.2,3.3;cout s(iarray,5)endl;cout s(darray,3)endl;5-10 / p144 习题5-10# include template void aggr( t (*data)c)for (int i=0; ir; i+)datai0=0;for (int j=0; jc; j+) datai0+=dataij;void main() / 依题意解题时主函数可以不必书写int a35=0,1,2,3,4,0,2,3,4,5,0,3,4,5,6; aggr(a);for (int i=0; i3; i+)cout endl;for (int j=0; j5; j+)cout aij ;cout endl;5-11 / p144 习题 5-11# include # include / 计算行列式的值函数。以下函数定义参阅1982年12/月第1版上海教育出版社出版/ 上海计算技术研究所编电子计算机算法手册第/295页算法template t valueof( t (*sq)size ) / 说明：在vc+6.0/下形参不允许书写成 sqsize形式int i,j,k,n;t p,t,u;t msize;for (i=0; isize; i+) u=0;for (j=0; ju) ) u=abs(sqij);if (u1.0e-18) return 0;mi=u;if (u!=1)for (j=0; jsize; j+)sqij/=u;p=1;for (k=0; ksize-1; k+)n=k;t=abs(sqkk);for (j=k+1; jsize; j+)if (tabs(sqkj) t=abs(sqkj);n=j;if (t1.0e-18) return 0;if (n!=k) p=-p;for (i=k; isize; i+) t=sqik;sqik=sqin;sqin=t;p*=sqkk;for (i=k+1; isize; i+) t=sqik/sqkk;for (j=k+1; jsize; j+)sqij-=sqkj*t;t=p*sqsize-1size-1;for (k=0; ksize; k+)t*=mk;return t;void main() / 依题意解题时主函数可以不必书写 # define maxsize 4double datamaxsizemaxsize;int i,j;/ 输入行列式cout请输入maxsize行maxsize列的行列式：endl;for (i=0; imaxsize; i+)for (j=0; jdataij;/ 输出行列式coutendlendlmaxsize”行maxsize列的行列式：endl; for (i=0; imaxsize; i+)cout endl;for (j=0; jmaxsize; j+)cout setw(10)dataij; cout endl;/ 调用函数valueof()计算行列式的值 cout endlmaxsize行maxsize列的行列式的值为：;cout valueof(data)endl; / 调用函数模板求行列式值习题六 （p187）6-3 / p192 习