第七次课上课程序及习题参考程序.docx

1、第七次课上课程序及习题参考程序第七次课上课程序复习：例1：选择法排序：#include int main(void) int arr10, i, j, temp, min; printf(Please input 10 numbers:n); for (i = 0; I 10; i+) /*数组的输入*/ scanf(%d, &arri); for (i = 0; i 10 - 1; i+) /*选择法从小到大排序*/ min = i; for(j = i + 1; j arrj) min = j; if (min != i) temp = arri; arri = arrmin; arrmi

2、n = temp; printf(The sorted numbers:n); for (i = 0; i 10; i+) /*数组的输出*/ printf(%4d, arri); return 0;例2：冒泡法排序#include int main(void) int arr10, i, j, temp; printf(Please input 10 numbers:n); for ( i = 0; i 10; i+) /*数组的输入*/ scanf(%d, &arri); for (i = 0; i 10 - 1; i+) /*冒泡法从小到大排序*/ for ( j = 0; j arrj

3、+1) temp = arrj; arrj = arrj+1; arrj+1 = temp; printf(The sorted numbers:n); for ( i = 0; i 10; i+) /*数组的输出*/ printf(%4d, arri); return 0;函数独立：记忆1：理解并记忆例1和例2选择法、冒泡法函数的声明、调用、定义例1：选择法排序：#include void select_sort(int arr, int n); /*选择法排序函数声明*/int main(void) int arr10, i; printf(Please input 10 numbers:

4、n); for (i = 0; i 10; i+) scanf(%d, &arri); select_sort(arr, 10); /*选择法排序函数调用*/ printf(The sorted numbers:n); for (i = 0; i 10; i+) printf(%4d, arri); printf(n); return 0;void select_sort(int arr, int n) /*选择法排序函数定义*/ int i, j, min, temp; for (i = 0; i n - 1; i+) min = i; for(j = i + 1; j arrj) min

5、= j; if (min != i) temp = arri; arri = arrmin; arrmin = temp; 例2：冒泡法排序#include void bubble_sort(int arr, int n); /*冒泡法排序函数声明*/int main(void) int arr10, i; printf(Please input 10 numbers:n); for ( i = 0; i 10; i+) scanf(%d, &arri); bubble_sort(arr, 10); /*冒泡法排序函数调用*/ printf(The sorted numbers:n); for

6、 ( i = 0; i 10; i+) printf(%4d, arri); printf(n); return 0;void bubble_sort(int arr, int n) /*冒泡法排序函数定义*/ int i, j, temp; for (i = 0; i n - 1; i+) for ( j = 0; j arrj+1) temp = arrj; arrj = arrj+1; arrj+1 = temp; 例3：二维数组举例58题码：章节：数组考查点：键盘输入技巧的掌握和二维数组的应用难度：2编写一程序P2-216.C实现以下功能求任意的一个mm矩阵的最大数及其所在的行列数，m

7、(2=m=20)及矩阵元素从键盘输入(只考虑int型)。编程可用素 材：printf(Pleaseinputm:).、printf(nPleaseinputarray:n).、 printf(nmax=.,i=.,j=.n.。程序的运行效果应类似地如图1所示，图1中的5和1456723546435689323428347934647957464678543619434634323556724846525是从键盘输入的内容。图1 程序运行效果示例分析结果：输入： m和arr值输出：最大值max及其行标maxi和列标maxj算法：初值max=arr00，maxi=0，maxj=0；将max与arr

8、数组中所有元素进行比较找出最大值并记下其行标和列标#include int main(void) /*定义变量*/ int arr2020; int i, j, m, max, maxi, maxj; /*输入数据*/ printf(Please input m: ); scanf(%d, &m); printf(nPlease input array:n); for (i = 0; i m; i+) /*i控制行*/ for (j = 0; j m; j+) /*j控制列*/ scanf(%d, &arrij); /*处理数据*/ max = arr00; maxi = 0; maxj =

9、0; for (i = 0; i m; i+) /*i控制行*/ for (j = 0; j m; j+) /*j控制列*/ if (max arrij) max = arrij; maxi = i; maxj = j; /*输出数据*/ printf(nmax=%d,i=%d,j=%dn, max, maxi, maxj); return 0;类似有59题（p2-217.c）、60题（p2-218.c）例4：顺序查找函数有序表和无序表皆可int search(int a, int n, int x) int i; for (i=0; in; i+) if (ai = x) return (i

10、); return (-1);例5：折半查找函数适用于有序表int binsearch(int a, int n, int x) int low, high, mid; low = 0; high = n - 1; while (low amid) low = mid + 1; else if (x amid) high = mid - 1; else return (mid); return(-1); 记忆2：排序+查找：理解并记忆例6例6：输入10个数，从小到大排序，然后输入一个数，要求找出该数是数组中第几个(序号从1开始计算)元素：#include void select_sort(in

11、t arr, int n); /*函数声明*/int binsearch(int a, int n, int x);/*函数声明*/ int main(void) int arr10, i, num, flag; printf(Please input 10 numbers:n); for (i = 0; i 10; i+) scanf(%d, &arri); select_sort(arr, 10); /*函数调用*/ printf(The sorted numbers:n); for (i = 0; i 10; i+) printf(%4d, arri); printf(n); print

12、f(input the num to look for:n); scanf(%d, &num); flag = binsearch(arr, 10, num);/*函数调用*/ if (flag != -1 ) printf(the position of %d is %dn, num, flag+1); else printf(not find!n); return 0;void select_sort(int arr, int n) /*函数定义*/ int i, j, min, temp; for (i = 0; i n - 1; i+) min = i; for(j = i + 1;

13、j arrj) min = j; if (min != i) temp = arri; arri = arrmin; arrmin = temp; int binsearch(int a, int n, int x) /*函数定义*/ int low, high, mid; low = 0; high = n - 1; while (low amid) low = mid + 1; else if (x amid) high = mid - 1; else return (mid); return(-1); 字符串与字符数组记忆3：理解并记忆例7和例8一维字符数组和二维字符数组的输入和输出例7

14、_1：一维字符数组逐个输入与输出，理解并记忆for中循环终止条件#include int main(void) int i; char str80, ch; printf(Input a string:); for (i=0; (ch = (char)getchar() != n; i+) /*字符串输入遇上n结束*/ stri = ch; stri = 0; /*人为加上结束符*/ printf(output a string:); /*字符串输出遇上结束符 0 结束*/ for (i=0; stri != 0; i+) putchar(stri); printf(n); return 0;

15、例7_2：一维字符数组整体输入使用scanf()和printf()不能输入带空格字符串使用gets()和puts()可输入带空格的字符串#include int main(void) char str80; printf(Input a string: ); gets(str); /* scanf(%s, str); */ printf(Output a string: ); puts(str); /* printf(%s,str); */ printf(n); return 0;例8:二维字符数组的输入和输出#include int main(void) char str580; print

16、f(Input a string: ); for (i=0;i5;i+)gets(str); printf(Output a string: ); for (i=0; i5; i+)puts(str); printf(n); return 0;例9：53题码：章节：数组考查点：字符数组的处理，循环和判断难度：2编写一程序P2-220.C实现以下功能从键盘读入一行字符(约定：字符数127字节)，统计及输出其中的字母、数字、空格和其他符号的个数。编程可用素 材：printf(Please input string:).、 printf(nzimu=.,shuzi=.,kongge=.,qita=.

17、。程序的运行效果应类似地如图1所示，图1中的gfaskl45623cvsac,53dwafl faf32535是从键盘输入的内容。#include int main(void) char str128; int i, zimu = 0, shuzi = 0, kongge = 0, qita = 0; printf(Please input string:); gets(str); for (i= 0; stri != 0; i+) if (stri = a) & (stri = A) & (stri = 0) & (stri = 9) shuzi+; else if (stri = ) ko

18、ngge+; else qita+; printf(nzimu=%d,shuzi=%d,kongge=%d,qita=%dn, zimu, shuzi, kongge, qita); return 0;记忆4：理解并记忆例10中二维字符数组的输入和输出及排序例10：95题码：章节：数组考查点：排序算法，二维字符数组及字符串处理函数的应用 难度：3编写一程序P7-719.C实现以下功能从键盘上输入5个字符串(约定：每个字符串中字符数80字节)，对其进行升序排序并输出。编程可用素材：printf(Input 5 strings:n).、printf(-n).。程序的运行效果应类似地如图1所示，图1

19、中的 hello my friend Are you ready? help!是从键盘输入的内容。#include #include void sel_sort(char s81, int n);int main(void) char str581; int i; printf(Input 5 strings:n); for (i = 0; i 5; i+) gets(stri); sel_sort(str, 5); printf(-n); for (i = 0; i 5; i+) puts(stri); return 0;void sel_sort(char s81, int n) int

20、i, j, min; char temp81; for (i = 0; i n - 1; i+) min = i; for (j = i + 1; j 0) min = j; if (min != i) strcpy(temp, si); strcpy(si, smin); strcpy(smin, temp); 习题参考程序习题7-5P7-771.c思路：找到一个单词将其放在temp128数组中，并用count统计其长度，若countmaxlen，则将temp128中的暂存单词复制到max128中，并用maxlen（maxlen=count）记下该单词长度。#include #include

21、 int main(void) char str128, temp128, max128; int i, j = 0, count = 0, maxlen = 0; printf(please input a string:n); gets(str); for (i = 0; stri != 0; i+) if (stri != ) tempj = stri; j+; count+; else if (j != 0 ) /*处理单词和单词之间有多个空格*/ tempj = 0; if (count maxlen) strcpy(max, temp); maxlen = count; j = 0

22、; count = 0; if (count maxlen) /*处理最后一个单词是最长单词*/ strcpy(max, temp); maxlen = count; printf(nmax_length of the string is: %dn, maxlen); return 0;习题7-6P7-705.c#include int main(void) char ch; printf(Please input the letter of someday: ); while (ch = (char)getchar() != Y) switch (ch) case M: printf(Mon

23、dayn); break; case W: printf(Wednesdayn); break; case F: printf(Fridayn); break; case T: ch = (char)getchar(); if (ch = u) printf(Tuesdayn); else if (ch = h) printf(Thrusdayn); else ungetc(ch, stdin); /*将ch中的字符还给stdin即键盘*/ printf(data errorn); break; case S: ch = (char)getchar(); if (ch = a) printf(Saturdayn); else if (ch = u)