本文主要是介绍谭浩强【C语言程序设计】第七章习题详解，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
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。两个整数由键盘输入。
最大公约数
🌵方法一：暴力求解
//穷举法 #include <stdio.h>int gcd(int a, int b) {if (a == 0)return b;else if (b == 0)return a;else if(a == b)return a;//或者bint gcd_ret = a > b ? b : a;while (gcd_ret > 1){if ((a % gcd_ret == 0) && (b % gcd_ret == 0))break;gcd_ret--;}return gcd_ret;}int main() {int a, b;printf("请输入两个整数：");scanf("%d%d", &a, &b);int gcd_ret = gcd(a, b);printf("最大公约数为：%d\n", gcd_ret);return 0; }🌵方法二：更相减损法
//相减法 int gcd(int a, int b) {if (a == 0)return b;else if (b == 0)return a;else if(a == b)return a;//或者bint gcd_ret = 0;while (a != b){gcd_ret = a > b ? (a -= b) : (b -= a);}return gcd_ret; }int main() {int a, b;printf("请输入两个整数：");scanf("%d%d", &a, &b);int gcd_ret = gcd(a, b);printf("最大公约数为：%d\n", gcd_ret);return 0; }🌵方法三：辗转相除法
//辗转相除法 int gcd(int a, int b) {if (a == 0)return b;else if (b == 0)return a;else if (a == b)return a;//或者bint mod = a % b;while (mod){a = b;b = mod;mod = a % b;}return b; }int main() {int a, b;printf("请输入两个整数：");scanf("%d%d", &a, &b);int gcd_ret = gcd(a, b);printf("最大公约数为：%d\n", gcd_ret);return 0; }🌵方法四：递归写法
//辗转相除法---递归实现 int gcd(int a, int b) {if (b == 0)return a;elsereturn gcd(b, a % b); }int main() {int a, b;printf("请输入两个整数：");scanf("%d%d", &a, &b);int gcd_ret = gcd(a, b);printf("最大公约数为：%d\n", gcd_ret);return 0; }最小公倍数
🌵方法一：暴力求解
//穷举法 int lcm(int a, int b) {if (a * b == 0)return 0;int lcm_ret = a > b ? a : b;while (1){if ((lcm_ret % a == 0) && (lcm_ret % b == 0))break;lcm_ret++;}return lcm_ret; }int main() {int a, b;printf("请输入两个整数：");scanf("%d%d", &a, &b);int lcm_ret = lcm(a, b);printf("最小公倍数为：%d\n", lcm_ret);return 0; }🌵方法二：公式法
//辗转相除法---递归实现 int gcd(int a, int b) {if (b == 0)return a;elsereturn gcd(b, a % b); }//公式法 lcm = a * b / gcd(a,b) int lcm(int a, int b) {if (a * b == 0)return 0;return a * b / gcd(a, b); }int main() {int a, b;printf("请输入两个整数：");scanf("%d%d", &a, &b);int gcd_ret = gcd(a, b);printf("最大公约数为：%d\n", gcd_ret);int lcm_ret = lcm(a, b);printf("最小公倍数为：%d\n", lcm_ret);return 0; }
2. 求方程ax²+bx+c=0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 disc;//判别式：disc = b^2-4ac float x1, x2;//两个根 float p, q;//两个虚根void greater_than_zero(int a, int b) {//x1 = -b + sqrt(disc) / 2a//x2 = -b - sqrt(disc) / 2ax1 = -b + sqrt(disc) / (2 * a);x2 = -b - sqrt(disc) / (2 * a); }void equal_to_zero(int a, int b) {//x1 = x2 = -b / 2ax1 = x2 = -b / (2 * a); }void less_than_zero(int a, int b) {//p = -b / 2a//q = sqrt(-disc) / 2ap = -b / 2 * a;q = sqrt(-disc) / (2 * a); }int main() {int a, b, c;printf("请输入a,b,c：");scanf("%d%d%d", &a, &b, &c);disc = b * b - 4 * a * c;//计算判别式if (disc > 0){greater_than_zero(a, b);printf("disc>0的两个根为：x1=%f，x2=%f\n", x1, x2);}else if (disc == 0){equal_to_zero(a, b);printf("disc=0的两个根为：x1=%f，x2=%f\n", x1, x2);}else//disc < 0{less_than_zero(a, b);printf("disc<0的两个根为：x1=%f，x2=%f\n", p+q, p-q);}return 0; }
3.写一个判素数的函数，在主函数输入一个整数，输出是否为素数的信息。
#include <stdio.h> #include <stdbool.h> #include <math.h>bool IsPrime(value) {for (int i = 2; i <= sqrt(value); ++i){if (value % i == 0)return false;}return true; }int main() {bool flag;int value;printf("请输入 value: ");scanf("%d", &value);flag = IsPrime(value);if (flag)printf("%d 是素数\n", value);elseprintf("%d 不是素数\n", value);return 0; }
4.写一个函数，使给定的一个3×3的二维整型数组转置，即行列互换。
#include <stdio.h>void PrintArray(int arr[3][3]) {for (int i = 0; i < 3; ++i){for (int j = 0; j < 3; ++j){printf("%d ", arr[i][j]);}printf("\n");} }void ReverseArray(int arr[3][3]) {for (int i = 0; i < 3; ++i){for (int j = 0; j < i; ++j){int tmp = arr[i][j];arr[i][j] = arr[j][i];arr[j][i] = tmp;}} }int main() {int arr[3][3] = { 1,2,3,4,5,6,7,8,9 };PrintArray(arr);printf("\n");ReverseArray(arr);PrintArray(arr);return 0; }
5.写一个函数，使输入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>void ReverseString(char str[]) {int start = 0;int end = strlen(str) - 1;while (start < end){char tmp = str[start];str[start] = str[end];str[end] = tmp;start++;end--;} }int main() {char str[100] = { 0 };printf("请输入字符串:>");gets(str);printf("原始字符串为：%s\n", str);ReverseString(str);printf("反序字符串为：%s\n", str);return 0; }
6.写一个函数，将两个字符串连接。
#include <stdio.h>void ConcatString(char dest[], char str1[], char str2[]) {int i = 0;while (str1[i] != '\0'){dest[i] = str1[i];i++;}int j = 0;while (str2[j] != '\0'){dest[i++] = str2[j++];}dest[i] = '\0'; }int main() {char str1[100] = { 0 };char str2[100] = { 0 };char str[200] = { 0 };printf("请输入第一个字符串:>");gets(str1);printf("请输入第二个字符串:>");gets(str2);ConcatString(str, str1, str2);printf("新字符串为：%s\n", str);return 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>void Copy(char dest[], char str[]) {int i = 0;int j = 0;while (str[i] != '\0'){if (str[i] == 'a' || str[i] == 'A' || str[i] == 'o' || str[i] == 'O' || str[i] == 'e' || str[i] == 'E' || str[i] == 'i' || str[i] == 'I' || str[i] == 'u' || str[i] == 'U'){dest[j++] = str[i];}i++;}dest[j] = '\0'; }int main() {char str[100] = { 0 };char dest[100] = { 0 };printf("请输入一个字符串:>");gets(str);printf("原始字符串为：%s\n", str);Copy(dest, str);printf("元音字符串为：%s\n", dest);return 0; }
8.写一个函数，输入一个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。如输入 1990,应输出“1 9 9 0”。
#include <stdio.h>void OutString(char digits[]) {int i = 0;while (digits[i] != '\0'){printf("%c", digits[i]);if (digits[i + 1] == '\0')break;printf(" ");i++;} }int main() {char digits[5] = { 0 };printf("请输入一个4位数字符串:>");scanf("%s", digits);OutString(digits);printf("\n");return 0; }
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，在主函数中输入字符串以及输出上述的结果。
#include <stdio.h>int letter, digit, space, others;void CountString(char str[]) {int i = 0;while (str[i] != '\0'){//统计字母if (str[i] >= 'a' && str[i] <= 'z' || str[i] >= 'A' && str[i] <= 'Z')letter++;else if (str[i] >= '0' && str[i] <= '9')//统计数字digit++;else if (str[i] == ' ')//统计空格space++;else//统计其它字符others++;i++;} }int main() {char str[256] = { 0 };printf("请输入一个字符串:>");gets(str);CountString(str);printf("字母个数：%d\n数字个数：%d\n空格个数：%d\n其它字符个数：%d\n",letter, digit, space, others);return 0; }
10.写一个函数，输入一行字符，将此字符串中最长的单词输出。
#include <stdio.h> #include <string.h>void FindLongWord(char str[], char word[]) {int i = 0, j = 0, len = 0;while (str[i] != '\0'){j = i;while (str[j] != ' ' && str[j] != '\0'){j++;}len = j - i;if (len > strlen(word)){strncpy(word, str + i, len);}j++;i = j;} }int main() {char str[256] = "A gentleman who speaks and recites is as gentle as jade";char word[256] = { 0 };FindLongWord(str, word);printf("最长单词为：%s\n", word);return 0; }
11.写一个函数，用“起泡法”对输入的10个字符按由小到大顺序排列。
#include <stdio.h> #include <string.h>void BubbleSort(char str[]) {int n = strlen(str);for (int i = 0; i < n - 1; i++){for (int j = 0; j < n - 1 - i; j++){if (str[j] > str[j + 1]){char tmp = str[j];str[j] = str[j + 1];str[j + 1] = tmp;}}} }int main() {char str[11] = { 0 };printf("请输入10个字符：");for (int i = 0; i < 10; i++){scanf("%c", &str[i]);}printf("原始字符序列为：%s\n", str);BubbleSort(str);printf("排序后字符序列为：%s\n", str);return 0; }
12. 用牛顿迭代法求根。方程为ax³+bx²+cx+d=0,系数a,b,c,d的值依次为1,2,3,4，由主函数输入。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>double Root(int a, int b, int c, int d, double x) {double x0;double f, f1;do{x0 = x;f = a * pow(x0, 3) + b * pow(x0, 2) + c * x0 + d;f1 = 3 * a * pow(x0, 2) + 2 * b * x0 + c;x = x0 - f / f1;} while (fabs(x - x0) >= 1e-3);return x; }int main() {int a, b, c, d;double x;printf("请输入系数a b c d x:>");scanf("%d%d%d%d%lf", &a, &b, &c, &d, &x);double res = Root(a, b, c, d, x);printf("root = %lf\n", res);return 0; }
13. 用递归方法求 n阶勒让德多项式的值。
#include <stdio.h>float Ploy(int n, int x) {if (n == 0)return 1;else if (n == 1)return x;return ((2 * n - 1) * x - Ploy(n - 1, x) - (n - 1) * Ploy(n - 2, x)) / n; }int main() {int n, x;printf("请输入n和x的值:>");scanf("%d%d", &n, &x);float ret = Ploy(n, x);printf("%d阶勒让德多项式的值为：%f", n, ret);return 0; }
14. 输入 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，Xi为某一学生的平均分。
#include <stdio.h> #define M 10 //学生人数 #define N 5 //课程数float avg_stu[M];//每个学生的平均分 float avg_course[N];//每门课程的平均分float hight;//最高分 int stu_index, course_index;//最高分的学生和课程成绩下标 float variance;//方差//计算每个学生的平均分 void avg_score_stu(float score[M][N]) {float sum = 0.0;for (int i = 0; i < M; i++){sum = 0.0;for (int j = 0; j < N; j++){sum += score[i][j];}avg_stu[i] = sum / N;} }//计算每门课程的平均分 void avg_score_course(float score[M][N]) {float sum = 0.0;for (int i = 0; i < N; i++){sum = 0.0;for (int j = 0; j < M; j++){sum += score[j][i];}avg_course[i] = sum / N;} }//所有分数中最高的分数所对应的学生和课程 float hight_course(float score[M][N]) {float hight = 0.0;for (int i = 0; i < M; i++){for (int j = 0; j < N; j++){if (score[i][j] > hight){hight = score[i][j];stu_index = i;course_index = j;}}}return hight; }//计算平均分方差 float variance_avg_score() {float sum_avg_square = 0.0;//平均分平方和float sum_avg_score = 0.0;//平均分和for (int i = 0; i < M; i++){sum_avg_square += avg_stu[i] * avg_stu[i];sum_avg_score += avg_stu[i];}return (sum_avg_square / M - (sum_avg_score / M) * (sum_avg_score / M)); }//打印 void print_result(float score[M][N]) {printf("N0 course1 course2 course3 course4 course5 average\n");for (int i = 0; i < M; i++){printf("N%-5d", i + 1);for (int j = 0; j < N; j++){printf("%-10.1f", score[i][j]);}//显示每个学生的平均分printf("%-10.1f\n", avg_stu[i]);}printf("%-5s ", "avg");for (int i = 0; i < N; i++){//显示每一门课的平均分printf("%-10.1f", avg_course[i]);}printf("\n");printf("最高分为：%.1f分，是第%d位同学的第%d门课程。\n", hight, stu_index+1, course_index+1);printf("平均分的方差为：%.1f\n", variance); }int main() {float score[M][N] ={{1,2,3,4,5},{2,3,4,5,6},{3,4,5,6,7},{4,5,6,7,8},{5,6,7,8,9},{6,7,8,3,1},{7,8,8,1,2},{8,8,1,2,3},{7,1,2,3,4},{4,3,2,1,8}};//计算每个学生的平均分avg_score_stu(score);//计算每门课程的平均分avg_score_course(score);//所有分数中最高的分数所对应的学生和课程hight = hight_course(score);//计算平均分方差variance = variance_avg_score(score);//打印print_result(score);return 0; }
15. 写几个函数:
①输入 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要求输入一个职工号，用折半查找法找出该职工的姓名，从主函数输入要查找的职工号，输出该职工姓名。
#include <stdio.h> #include <string.h>#define N 10 #define NAME_SIZE 10//输入10个职工的姓名和职工号 void Input(int id[], char name[][NAME_SIZE]) {for (int i = 0; i < N; i++){printf("输入职工号：");scanf("%d", &id[i]);getchar(); //忽略输入的回车printf("输入职工的姓名：");gets(name[i]);} }//在屏幕上输出 void Output(int id[], char name[][NAME_SIZE]) {for (int i = 0; i < N; i++){printf("[%d] : [%s]\n", id[i], name[i]);} }//排序 void Sort(int id[], char name[][NAME_SIZE]) {char tmp_name[NAME_SIZE];for (int i = 0; i < N - 1; i++){for (int j = 0; j < N - 1 - i; j++){if (id[j] > id[j + 1]){int tmp_id = id[j];strcpy(tmp_name, name[j]);id[j] = id[j + 1];strcpy(name[j], name[j + 1]);id[j + 1] = tmp_id;strcpy(name[j + 1], tmp_name);}}} }//折半法查找职工号 void Search(int id[], char name[][NAME_SIZE], int key) {int low = 0;int high = N - 1;int mid;while (low <= high){mid = (low + high) / 2;if (key == id[mid])break;if (key < id[mid])high = mid - 1;elselow = mid + 1;}if (low <= high)printf("职工号为%d的职工姓名为：%s\n", key, name[mid]);elseprintf("要查找的职工号为%d的职工不存在。\n", key); }int main() {int id[N];//职工号char name[N][NAME_SIZE];//职工号对应的职工姓名Input(id, name);Output(id, name);Sort(id, name);printf("\n");Output(id, name);int no;while (1){printf("请输入要查找的职工号：");scanf("%d", &no);Search(id, name, no);}return 0; }
16.写一个函数，输入一个十六进制数，输出相应的十进制数。
#include <stdio.h>unsigned int HextoDec(char hex[]) {unsigned int ret = 0;int i = 0;while (hex[i] != '\0'){if (hex[i] >= 'a' && hex[i] <= 'z')ret = 16 * ret + hex[i] - 'a' + 10;else if (hex[i] >= 'A' && hex[i] <= 'Z')ret = 16 * ret + hex[i] - 'A' + 10;elseret = 16 * ret + hex[i] - '0';i++;}return ret; }int main() {char hex[9] = { 0 };// FF FF FF FFprintf("请输入一个十六进制数:>");scanf("%s", hex);unsigned int ret = HextoDec(hex);printf("0x%s ==> %u", hex, ret);return 0; }
17.用递归法将一个整数n转换成字符串。例如，输入 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>void Convert(int n) {if (n / 10 != 0)Convert(n / 10);printf("%c", n % 10 + '0'); }int main() {int num = 0;printf("请输入一个整数:>");scanf("%d", &num);printf("整数%d转换成字符串为: ", num);Convert(num);printf("\n");return 0; }
18. 给出年、月、日，计算该日是该年的第几天。
#include <stdio.h> #include <stdbool.h>bool IsLeap(int year) {return ((year % 4 == 0) && (year % 100 != 0) || (year % 400 == 0)); }int GetDayByYM(int year, int month) {//1 2 3 4 5 6 7 8 9 10 11 12int days[13] = { 29, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };if ((month == 2) && IsLeap(year))return days[0];return days[month]; }int GetDayByYMD(int year, int month, int day) {int days = 0;for (int i = 1; i < month; i++){days += GetDayByYM(year, i);}return days + day; }int main() {int year, month, day;printf("请输入年 月 日:>");scanf("%d%d%d", &year, &month, &day);int days = GetDayByYMD(year, month, day);printf("%d年%d月%d日是该年的第%d天\n", year, month, day, days);return 0; }
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