做设计时可以参考的网站,wordpress后台如何登录,网站平台建设的实训报告,厦门网站建设一般多少钱C程序设计(第5版)谭浩强习题解答
第7章 用函数实现模块化程序设计
1.写两个函数,分别求两个整数的最大公约数和最小公倍数,用主函数调用这两个函数,并输出结果。两个整数由键盘输人。
//7.1.1
#include stdio.h
int main()
{int hcf(int, int);int lcd(int, int, in…C程序设计(第5版)谭浩强习题解答
第7章 用函数实现模块化程序设计
1.写两个函数,分别求两个整数的最大公约数和最小公倍数,用主函数调用这两个函数,并输出结果。两个整数由键盘输人。
//7.1.1
#include stdio.h
int main()
{int hcf(int, int);int lcd(int, int, int);int u, v, h, l;scanf(%d,%d, u, v);h hcf(u, v);printf(H.C.F%d\n, h);l lcd(u, v, h);printf(L.C.D%d\n, l);return 0;
}int hcf(int u, int v)
{int t, r;if (v u){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);
}//7.1.2
#include stdio.h
int Hcf, Lcd;
int main()
{void hcf(int, int);void lcd(int, int);int u, v;scanf(%d,%d, u, v);hcf(u, v);lcd(u, v);printf(H.C.F%d\n, Hcf);printf(L.C.D%d\n, Lcd);return 0;
}void hcf(int u, int v)
{int t, r;if (v u){t u; u v; v t;}while ((r u % v) ! 0){u v;v r;}Hcf v;
}void lcd(int u, int v)
{Lcd u * v / Hcf;
}2.求方程 ax2bxc0的根,用3个函数分别求当:b2−4ac大于0、等于0和小于0时的根并输出结果。从主函数输入a,b,c的值。
#include stdio.h
#include math.h
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;printf(input a,b,c:);scanf(%f,%f,%f, a, b, c);printf(equation: %5.2f*x*x%5.2f*x%5.2f0\n, a, b, c);disc b * b - 4 * a*c;printf(root:\n);if (disc 0){greater_than_zero(a, b);printf(x1%f\t\tx2%f\n, x1, x2);}else if (disc 0){equal_to_zero(a, b);printf(x1%f\t\tx2%f\n, x1, x2);}else{smaller_than_zero(a, b);printf(x1%f%fi\tx2%f-%fi\n, p, q, p, q);}return 0;
}void greater_than_zero(float a, float b)
{x1 (-b sqrt(disc)) / (2 * a);x2 (-b - sqrt(disc)) / (2 * a);
}void equal_to_zero(float a, float b)
{x1 x2 (-b) / (2 * a);
}void smaller_than_zero(float a, float b)
{p -b / (2 * a);q sqrt(-disc) / (2 * a);
}3.写一个判素数的函数,在主函数输人一个整数,输出是否为素数的信息。
#include stdio.h
int main()
{int prime(int);int n;printf(input an integer:);scanf(%d, n);if (prime(n))printf(%d is a prime.\n, n);elseprintf(%d is not a prime.\n, n);return 0;
}int prime(int n)
{int flag 1, i;for (i 2; i n / 2 flag 1; i)if (n%i 0)flag 0;return(flag);
}4.写一个函数,使给定的一个3X3的二维整型数组转置,即行列互换。
#include stdio.h
#define N 3
int array[N][N];
int main()
{void convert(int array[][3]);int i, j;printf(input array:\n);for (i 0; i N; i)for (j 0; j N; j)scanf(%d, array[i][j]);printf(\noriginal array :\n);for (i 0; i N; i){for (j 0; j N; j)printf(%5d, array[i][j]);printf(\n);}convert(array);printf(convert array:\n);for (i 0; i N; i){for (j 0; j N; j)printf(%5d, array[i][j]);printf(\n);}return 0;
}void convert(int array[][3])
{int i, j, t;for (i 0; i N; i)for (j i 1; j N; j){t array[i][j];array[i][j] array[j][i];array[j][i] t;}
}5.写一个函数,使输人的一个字符串按反序存放,在主函数中输入和输出字符串。
#include stdio.h
#include string.h
int main()
{void inverse(char str[]);char str[100];printf(input string:);scanf(%s, str);inverse(str);printf(inverse string:%s\n, str);return 0;
}void inverse(char str[])
{char t;int i, j;for (i 0, j strlen(str); i (strlen(str) / 2); i, j--){t str[i];str[i] str[j - 1];str[j - 1] t;}
}6.写一个函数,将两个字符串连接。
#include stdio.h
int main()
{void concatenate(char string1[], char string2[], char string[]);char s1[100], s2[100], s[100];printf(input string1:);scanf(%s, s1);printf(input string2:);scanf(%s, s2);concatenate(s1, s2, s);printf(\nThe new string is %s\n, s);return 0;
}void concatenate(char string1[], char string2[], char string[])
{int i, j;for (i 0; string1[i] ! \0; i)string[i] string1[i];for (j 0; string2[j] ! \0; j)string[i j] string2[j];string[i j] \0;
}7.写一个函数,将一个字符串中的元音字母复制到另一字符串,然后输出。
#include stdio.h
int main()
{void cpy(char[], char[]);char str[80], c[80];printf(input string:);gets(str);cpy(str, c);printf(The vowel letters are:%s\n, c);return 0;
}void cpy(char s[], char c[])
{int i, j;for (i 0, j 0; s[i] ! \0; i)if (s[i] a || s[i] A || s[i] e || s[i] E || s[i] i ||s[i] I || s[i] o || s[i] O || s[i] u || s[i] U){c[j] s[i];j;}c[j] \0;
}8.写一个函数,输人一个4位数字要求输出这4个数字字符,但每两个数字间空一个空格。如输人1990,应输出“1 9 9 0”。
#include stdio.h
#include string.h
int main()
{char str[80];void insert(char[]);printf(input four digits:);scanf(%s, str);insert(str);return 0;
}void insert(char str[])
{int i;for (i strlen(str); i 0; i--){str[2 * i] str[i];str[2 * i - 1] ;}printf(output:\n%s\n, str);
}9.编写一个函数,由实参传来一个字符串,统计此字符串中字母、数字、空格和其他字符的个数,在主函数中输人字符串以及输出上述的结果。
#include stdio.h
int letter, digit, space, others;
int main()
{void count(char[]);char text[80];printf(input string:\n);gets(text);printf(string:);puts(text);letter 0;digit 0;space 0;others 0;count(text);printf(\nletter:%d\ndigit:%d\nspace:%d\nothers:%d\n, letter, digit, space, others);return 0;
}void count(char str[])
{int i;for (i 0; str[i] ! \0; i)if ((str[i] a str[i] z) || (str[i] A str[i] Z))letter;else if (str[i] 0 str[i] 9)digit;else if (str[i] 32)space;elseothers;
}10.写一个函数,输人一行字符,将此字符串中最长的单词输出。
#include stdio.h
#include string.h
int main()
{int alphabetic(char);int longest(char[]);int i;char line[100];printf(input one line:\n);gets(line);printf(The longest word is :);for (i longest(line); alphabetic(line[i]); i)printf(%c, line[i]);printf(\n);return 0;
}int alphabetic(char c)
{if ((c a c z) || (c Ac z))return(1);elsereturn(0);
}int longest(char string[])
{int len 0, i, length 0, flag 1, place 0, point;for (i 0; i strlen(string); i)if (alphabetic(string[i]))if (flag){point i;flag 0;}elselen;else{flag 1;if (len length){length len;place point;len 0;}}return(place);
}11.写一个函数,用“起泡法”对输人的10个字符按由小到大顺序排列。
#include stdio.h
#include string.h
#define N 10
char str[N];
int main()
{void sort(char[]);int i, flag;for (flag 1; flag 1;){printf(input string:\n);scanf(%s, str);if (strlen(str) N)printf(string too long,input again!);elseflag 0;}sort(str);printf(string sorted:\n);for (i 0; i N; i)printf(%c, str[i]);printf(\n);return 0;
}void sort(char str[])
{int i, j;char t;for (j 1; j N; j)for (i 0; (i N - j) (str[i] ! \0); i)if (str[i] str[i 1]){t str[i];str[i] str[i 1];str[i 1] t;}
}12.用牛顿迭代法求根。方程为ax3bx2cxd0,系数a,b,c,d的值依次为1,2,3,4,由主函数输人。求x在1附近的一个实根。求出根后由主函数输出。
#include stdio.h
#include math.h
int main()
{float solut(float a, float b, float c, float d);float a, b, c, d;printf(input a,b,c,d:);scanf(%f,%f,%f,%f, a, b, c, d);printf(x%10.7f\n, solut(a, b, c, d));return 0;
}float solut(float a, float b, float c, float d)
{float x 1, x0, f, f1;do{x0 x;f ((a*x0 b)*x0 c)*x0 d;f1 (3 * a*x0 2 * b)*x0 c;x x0 - f / f1;} while (fabs(x - x0) 1e-3);return(x);
}13.用递归方法求n阶勒让德多项式的值,递归公式为
#include stdio.h
#define N 10
#define M 5
float score[N][M];
float a_stu[N], a_cour[M];
int r, c;int main()
{int i, j;float h;float s_var(void);float highest();void input_stu(void);void aver_stu(void);void aver_cour(void);input_stu();aver_stu();aver_cour();printf(\n NO. cour1 cour2 cour3 cour4 cour5 aver\n);for (i 0; i N; i){printf(\n NO %2d , i 1);for (j 0; j M; j)printf(%8.2f, score[i][j]);printf(%8.2f\n, a_stu[i]);}printf(\naverage:);for (j 0; j M; j)printf(%8.2f, a_cour[j]);printf(\n);h highest();printf(highest:%7.2f NO. %2d course %2d\n, h, r, c);printf(variance %8.2f\n, s_var());return 0;
}void input_stu(void)
{int i, j;for (i 0; i N; i){printf(\ninput score of student%2d:\n, i 1);for (j 0; j M; j)scanf(%f, score[i][j]);}
}void aver_stu(void)
{int i, j;float s;for (i 0; i N; i){for (j 0, s 0; j M; j)s score[i][j];a_stu[i] s / 5.0;}
}void aver_cour(void)
{int i, j;float s;for (j 0; j M; j){s 0;for (i 0; i N; i)s score[i][j];a_cour[j] s / (float)N;}
}float highest()
{float high;int i, j;high score[0][0];for (i 0; i N; i)for (j 0; j M; j)if (score[i][j] high){high score[i][j];r i 1;c j 1;}return(high);
}float s_var(void)
{int i;float sumx, sumxn;sumx 0.0;sumxn 0.0;for (i 0; i N; i){sumx a_stu[i] * a_stu[i];sumxn a_stu[i];}return(sumx / N - (sumxn / N)*(sumxn / N));
}14.输人10个学生5门课的成绩,分别用函数实现下列功能:
①计算每个学生的平均分;
②计算每门课的平均分;
③找出所有50个分数中最高的分数所对应的学生和课程;
④计算平均分方差:
其中,x;为某一学生的平均分。
#include stdio.h
#define N 10
#define M 5
float score[N][M];
float a_stu[N], a_cour[M];
int r, c;int main()
{int i, j;float h;float s_var(void);float highest();void input_stu(void);void aver_stu(void);void aver_cour(void);input_stu();aver_stu();aver_cour();printf(\n NO. cour1 cour2 cour3 cour4 cour5 aver\n);for (i 0; i N; i){printf(\n NO %2d , i 1);for (j 0; j M; j)printf(%8.2f, score[i][j]);printf(%8.2f\n, a_stu[i]);}printf(\naverage:);for (j 0; j M; j)printf(%8.2f, a_cour[j]);printf(\n);h highest();printf(highest:%7.2f NO. %2d course %2d\n, h, r, c);printf(variance %8.2f\n, s_var());return 0;
}void input_stu(void)
{int i, j;for (i 0; i N; i){printf(\ninput score of student%2d:\n, i 1);for (j 0; j M; j)scanf(%f, score[i][j]);}
}void aver_stu(void)
{int i, j;float s;for (i 0; i N; i){for (j 0, s 0; j M; j)s score[i][j];a_stu[i] s / 5.0;}
}void aver_cour(void)
{int i, j;float s;for (j 0; j M; j){s 0;for (i 0; i N; i)s score[i][j];a_cour[j] s / (float)N;}
}float highest()
{float high;int i, j;high score[0][0];for (i 0; i N; i)for (j 0; j M; j)if (score[i][j] high){high score[i][j];r i 1;c j 1;}return(high);
}float s_var(void)
{int i;float sumx, sumxn;sumx 0.0;sumxn 0.0;for (i 0; i N; i){sumx a_stu[i] * a_stu[i];sumxn a_stu[i];}return(sumx / N - (sumxn / N)*(sumxn / N));
}15.写几个函数:
①输人10个职工的姓名和职工号;
②按职工号由小到大顺序排序,姓名顺序也随之调整;
③要求输人一个职工号,用折半查找法找出该职工的姓名,从主函数输人要查找的职工号,输出该职工姓名。
#include stdio.h
#include string.h
#define N 10
int main()
{void input(int[], char name[][8]);void sort(int[], char name[][8]);void search(int, int[], char name[][8]);int num[N], number, flag 1, c;char name[N][8];input(num, name);sort(num, name);while (flag 1){printf(\ninput number to look for:);scanf(%d, number);search(number, num, name);printf(continue ot not(Y/N)?);getchar();c getchar();if (c N || c n)flag 0;}return 0;
}void input(int num[], char name[N][8])
{int i;for (i 0; i N; i){printf(input NO.: );scanf(%d, num[i]);printf(input name: );getchar();gets(name[i]);}
}void sort(int num[], char name[N][8])
{int i, j, min, templ;char temp2[8];for (i 0; i N - 1; i){min i;for (j i; j N; j)if (num[min] num[j]) min j;templ num[i];strcpy(temp2, name[i]);num[i] num[min];strcpy(name[i], name[min]);num[min] templ;strcpy(name[min], temp2);}printf(\n result:\n);for (i 0; i N; i)printf(\n %5d%10s, num[i], name[i]);
}void search(int n, int num[], char name[N][8])
{int top, bott, mid, loca, sign;top 0;bott N - 1;loca 0;sign 1;if ((n num[0]) || (n num[N - 1]))loca -1;while ((sign 1) (top bott)){mid (bott top) / 2;if (n num[mid]){loca mid;printf(NO. %d , his name is %s.\n, n, name[loca]);sign -1;}else if (n num[mid])bott mid - 1;elsetop mid 1;}if (sign 1 || loca -1)printf(%d not been found.\n, n);
}16.写一个函数,输人一个十六进制数,输出相应的十进制数。
#include stdio.h
#define MAX 1000
int main()
{int htoi(char s[]);int c, i, flag, flag1;char t[MAX];i 0;flag 0;flag1 1;printf(input a HEX number:);while ((c getchar()) ! \0 i MAX flag1){if (c 0 c 9 || c a c f || c A c F){flag 1;t[i] c;}else if (flag){t[i] \0;printf(decimal number %d\n, htoi(t));printf(continue or not?);c getchar();if (c N || c n)flag1 0;else{flag 0;i 0;printf(\ninput a HEX number:);}}}return 0;
}int htoi(char s[])
{int i, n;n 0;for (i 0; s[i] ! \0; i){if (s[i] 0 s[i] 9)n n * 16 s[i] - 0;if (s[i] a s[i] f)n n * 16 s[i] - a 10;if (s[i] A s[i] F)n n * 16 s[i] - A 10;}return(n);
}17.用递归法将一个整数n转换成字符串。例如输人483,应输出字符串”483”。n的位数不确定,可以是任意位数的整数。
#include stdio.h
int main()
{void convert(int n);int number;printf(input an integer: );scanf(%d, number);printf(output: );if (number 0){putchar(-); putchar( ); //先输出一个‘-’号和空格 number -number;}convert(number);printf(\n);return 0;
}void convert(int n)
{int i;if ((i n / 10) ! 0)convert(i);putchar(n % 10 0);putchar(32);
}18.给出年、月、日,计算该日是该年的第几天。
#include stdio.h
int main()
{int sum_day(int month, int day);int leap(int year);int year, month, day, days;printf(input date(year,month,day):);scanf(%d,%d,%d, year, month, day);printf(%d/%d/%d , year, month, day);days sum_day(month, day); //调用函数sum_day if (leap(year) month 3) //调用函数leap days days 1;printf(is the %dth day in this year.\n, days);return 0;
}int sum_day(int month, int day) //函数sum_day:计算日期
{int day_tab[13] { 0,31,28,31,30,31,30,31,31,30,31,30,31 };int i;for (i 1; i month; i)day day_tab[i]; //累加所在月之前天数 return(day);
} //函数leap:判断是否为闰年 int leap(int year)
{int leap;leap year % 4 0 year % 100 ! 0 || year % 400 0;return(leap);
}
