经典C语言源代码.doc

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1、【精品文档】如有侵权，请联系网站删除，仅供学习与交流经典C语言源代码.精品文档.经典C语言源代码1、(1)某年某月某日是星期几#includeint main()int year, month, day;while (scanf_s(%d%d%d, &year, &month, &day) != EOF)if (month = 1 | month = 2)/判断month是否为1或2year-;month += 12;int c = year / 100;int y = year - c * 100;int week = (c / 4) - 2 * c + (y + y / 4) + (13

2、* (month + 1) / 5) + day - 1;while (week0) week += 7; week %= 7;switch (week)case 1:printf(Mondayn); break;case 2:printf(Tuesdayn); break;case 3:printf(Wednesdayn); break;case 4:printf(Thursdayn); break;case 5:printf(Fridayn); break;case 6:printf(Saturdayn); break;case 0:printf(Sundayn); break;retur

3、n 0;1、(2)某年某月某日是第几天（一维数组）#include stdio.hvoid main() int i, flag, year, month, day, dayth;int month_day = 0,31,28,31,30,31,30,31,31,30,31,30,31 ;printf(请输入年/月/日：n);scanf_s(%d/%d/%d, &year, &month, &day);dayth = day;flag = (year % 400 = 0) | (year % 4 = 0 & year % 100 != 0);if (flag)month_day2 = 29;f

4、or (i = 1; i month; i+)dayth = dayth + month_dayi;printf(%d/%d/%d是第%d天n, year, month, day, dayth);2、30个数中找最小的数及其位置#include stdio.h# define SIZE 30void main() int i;float dataSIZE;int min;printf(请输入%d个浮点数：n,SIZE);for (i = 0; i SIZE; i+) /scanf_s(%f, &datai);datai = rand() % 30 + 1;printf(%f、, datai);

5、min = 0;for (i = 1; i SIZE; i+) if (datai datamin)min = i;printf(最小值是%5.2f,位置是%5dn, datamin, min);3、30个数从小到大排序（1）#include stdio.h# define SIZE 30void main() int i,j;float dataSIZE,temp;int min;printf(请输入%d个整型数：n,SIZE);for (i = 0; i SIZE; i+) scanf_s(%f, &datai);for (i = 0; i SIZE; i+) min = i;for (j

6、 = i + 1; j SIZE; j+)if (dataj datamin)min = j;temp = datamin;datamin = datai;datai = temp;printf(n排序后的结果是：n);for (i = 0; i SIZE; i+)printf(%5.2f, datai);（2）模块化程序（数组名作为函数参数）#include stdio.h# define SIZE 5void accept_array(float a, int size);void sort(float a, int size);void show_array(float a, int s

7、ize);void main() float scoreSIZE;accept_array(score, SIZE);printf(排序前：);show_array(score, SIZE);sort(score, SIZE);printf(排序后：);show_array(score, SIZE);void accept_array(float a, int size) int i;printf(请输入%d个分数：, size);for (i = 0; i size; i+)scanf_s(%f, &ai);void show_array(float a, int size) int i;f

8、or (i = 0; i size; i+)printf( %5.2f, ai);printf(n);void sort(float a,int size) int i, min, j;float temp;for (i = 0; i SIZE; i+) min = i;for (j = i + 1; j SIZE; j+)if (aj amin)min = j;temp = amin;amin = ai;ai = temp;4、（1）指针加减：#include stdio.h#define SIZE 10void main() int aSIZE = 1,2,3,4,5,6,7,8,9,10

9、 ;int *pa, i;pa = &a0; /pa=a;printf(n);for (i = 0; i SIZE; i+) printf( %d, *pa);/printf( %d, *(pa+1);pa+;（2）指针比较：#include stdio.h#define SIZE 10void main() int aSIZE = 1,2,3,4,5,6,7,8,9,10 ;int *pa, i;int *qa;pa = qa = &a0;printf(请输入%d整型数：,SIZE);for (; pa qa + SIZE; pa+)scanf_s(%d, pa);for (pa-; qa

10、= pa; pa-)printf( %d, *pa);5、两字符串相连：#include stdio.h#include string.hvoid str_cat(char str1, char str2);void main() int i, j;char str1160;char str280;printf(请输入第一个字符串：);gets(str1);printf(请输入第二个字符串：);gets(str2);str_cat(str1, str2);puts(str1);void str_cat(char str1, char str2) int i, j;i = 0;while (st

11、r1i != 0)i+;j = 0;while (str2j != 0) str1i = str2j;i+; j+;str1i = 0;6、二维数组（a，b转置）#include stdio.hvoid main() int i, j, b23;int a32 = 1,2,3,4,5,6 ;for (i = 0; i 2; i+) for (j = 0; j 3; j+)bij = aji;printf(na:n);for (i = 0; i 3; i+) for (j = 0; j 2; j+)printf(%5d, aij);printf(n);printf(nb:n);for(i = 0

12、; i 2; i+) for (j = 0; j 3; j+)printf(%5d, bij);printf(n);7、输入一个二维数组并输出（指针）#include stdio.hvoid main() int x23;int i, j;for (i = 0; i 2; i+)for (j = 0; j 3; j+)scanf_s(%d, *(x + i) + j);putchar(n);for (i = 0; i 2; i+)for (j = 0; j 3; j+)printf(%d , *(*(x + i) + j);putchar(n);8、冒泡法排序一个数组#include stdi

13、o.h#define size 10void maopao(int a);void main() int a10;int i;printf(请输入10个整数：n);for (i = 0; i 10; i+)scanf_s(%d, &ai);maopao(a);void maopao(int a) int i, j, temp;for (i = 0; i 9; i+) /进行9轮排序for (j = 0; j aj + 1)temp = aj;aj = aj + 1;/大的沉底，小的上浮aj + 1 = temp;printf(排序结果：n);for (i = 0; i 10; i+)print

14、f(%4d, ai);9、 两数组A，B，要求AB，如A：4，7，9B：1，3，5，8，9变换后A：1，3，5B：4，7，8，9，9#include void ReArranger(int* A, int* B, int m, int n) /A和B是各有m个和n个整数的非降序数组，本算法将B数组元素逐个插入到A中，使A中各元素均不大于B中各元素，且两数组仍保持非降序排列。int x, j, i;while (Am - 1B0)x = Am - 1;Am - 1 = B0; /交换Am-1和B0j = 1;while (jn & Bj= 0 & Aix)Ai + 1 = Ai-; /寻找B0的

15、插入位置Ai + 1 = x;void main() /这里主要介绍算法思想，主函数就简单写了int A3, B5, i;printf(输入第一个数组：);for (i = 0; i3; i+) scanf_s(%d, &Ai);printf(n输入第二个数组：);for (i = 0; i5; i+) scanf_s(%d, &Bi);ReArranger(A, B, 3, 5);printf(n输出第一个数组：);for (i = 0; i3; i+)printf(%d , Ai);printf(nn);printf(输出第二个数组：);for (i = 0; i5; i+)printf(

16、%d , Bi);printf(n);10、符合1+6+3=3+2+5=1+4+5有哪几组 A 1 B C 6 4D E F到 3 2 5#include void main() int a, b, c, d, e, f;for (a = 1; a = 6; a+)for (b = 1; b = 6; b+) if (b = a)continue;for (c = 1; c = 6; c+) if (c = a) | (c = b)continue;for (d = 1; d = 6; d+) if (d = a) | (d = b) | (d = c)continue;for (e = 1;

17、 e = 6; e+) if (e = a) | (e = b) | (e = c) | (e = d)continue;f = 21 - (a + b + c + d + e);if (a + b + d = d + e + f) & (a + b + d = a + c + f) printf( %dn, a);printf(%d %dn, b, c);printf(%d %d %dn, d, e, f);11、输入一串字符，升序排序，折半查找其中一字符#include void *sortString(char unsort, int length) for (int i = 0; i

18、length; i+) for (int j = i + 1; j unsortj) int temp = unsortj;unsortj = unsorti;unsorti = temp;void main() char s150;gets(s1);char value;scanf_s(%c, &value);printf(s1：%sn, s1);printf(value:%cn, value);int length = strlen(s1);printf(length:%dn, length);sortString(s1, length);printf(s1：%sn, s1);int st

19、art = 0, end = length - 1;int mid = (end + start) / 2;while (start s1mid)start = mid+1;elseend = mid-1;if (mid)printf(该字符在已知字符串中，即在第%d个n,mid);elseprintf(该字符不在已知字符串中n);12、100-300和500-700直接素数，m不被2到根号m直接任一整除#includeint isprime(int n)if (n2) return 0;for (int i = 2; in / 2; i+)if (n%i = 0) return 0;retu

20、rn 1;void main()int i, k = 0;for (i = 100; i = 300; i+)if (isprime(i)printf(%3d , i);k+;if (k % 10 = 0) printf(n);for (i = 500; i = 700; i+)if (isprime(i)printf(%3d , i);k+;if (k % 10 = 0) printf(n);printf(n);13、判断一个数是否是素数#include#includevoid main()int m, i, k;printf(请输入一个整数：);scanf_s(%d, &m);k = (i

21、nt)sqrt(m);for (i = 2; i k)printf(%d 是素数。n, m);elseprintf(%d 不是素数。n, m);14、一个数是否含有数字5#include #include bool is5Num(int num) int temp = num % 10;while (temp != 5 & num 10) num = num / 10;temp = num % 10;if (temp != 5)return false;elsereturn true;void main() int num;printf(输入一个数：n);scanf_s(%d, &num);i

22、f (is5Num(num)printf(含5n);elseprintf(不含5n);15、一个排好序的数组，插入一个数#include void main() int a11 = 1,2,3,4,5,6,7,8,9,10 ;int num;printf(插入前数组为：n);for (int i = 0; i a9)a10 = num;else for (int i = 0; i 10;i+)if (num = i; j-)aj + 1 = aj;ai = num;break;printf(插入后数组为：n);for (int i = 0; i = 10; i+)printf(%3d, ai)

23、;printf(n);16、牛顿迭代法：#include #include double func(double x) /函数return x*x*x + 2.0*x*x + 3.0*x + 4.0;double func1(double x) /导函数return 3 * x*x + 4 * x + 3;void Newton(double x0,double precision)/迭代次数double x1;int k;if (func1(x0) = 0.0) /若通过初值，函数返回为0printf(迭代过程中倒数为0！n);return;x1 = x0 - func(x0) / func

24、1(x0);/进行牛顿迭代计算while (!(func1(x1 - x0) precision | fabs(func(x1) precision) x0 = x1;/准备下一次迭代if (func1(x0) = 0.0)/若通过初值，函数返回值为0printf(迭代过程中倒数为0！n);x1 = x0 - func(x0) / func1(x0);/进行牛顿迭代计算void main() double x, precision;printf(输入初始迭代值x0：n);scanf_s(%lf, &x);printf(迭代要求的精度：n);scanf_s(%lf, &precision);Ne

25、wton(x, precision);/若函数返回值为1printf(该值附近的跟为：%lfn, x);getchar();getchar();17、起始时间到终止时间天数#include stdio.hvoid main() int start3, end3;printf(请输入开始日期，如1964.2.19：n);scanf_s(%d.%d.%d, &start0, &start1, &start2);printf(请输入结束日期，如2001.10.20：n);scanf_s(%d.%d.%d, &end0, &end1, &end2);int sum = 0;for (int mid =

26、 start0; mid end0; mid+) if (mid % 400 = 0) | (mid % 4 = 0 & mid % 100 != 0) sum = sum + 366;elsesum = sum + 365;sum = sum - indexday(start0,start1,start2) + indexday(end0,end1,end2);printf(在%d.%d.%d-%d.%d.%d之间有%d天n, start0,start1,start2,end0,end1,end2, sum);int indexday(int year, int month, int day

27、) int i, flag, dayth;int month_day = 0,31,28,31,30,31,30,31,31,30,31,30,31 ;dayth = day;flag = (year % 400 = 0) | (year % 4 = 0 & year % 100 != 0);if (flag)month_day2 = 29;for (i = 1; i month; i+)dayth = dayth + month_dayi;return dayth;18、递归求1*1+2*2+3*3+n*n#include stdio.hlong Element(int n) if (n =

28、 1)return 1 * 1;elsereturn Element(n - 1) + n*n;void main() int n;printf(请输入n的值：n);scanf_s(%d, &n);printf(所求值为%dn, Element(n);19、最大公约数（辗转相除）#include void main() /* 辗转相除法求最大公约数 */int m, n, a, b, t, c;printf(Input two integer numbers:n);scanf_s(%d%d, &a, &b);m = a; n = b;while (b != 0) /* 余数不为0，继续相除，直

29、到余数为0 */c = a%b; a = b; b = c;printf(The largest common divisor:%dn, a);printf(The least common multiple:%dn, m*n / a);20、杨辉三角#include void main()int i, j, n, k;printf(Enter n:); scanf_s(%d, &n);for (i = 1; i = n; i+)k = 1;for (j = 1; ji; j+)printf(%3d, k);k = k*(i - j) / j;/每次要打印的下一个数等于前一个数乘以其所在行数和

30、列数的差再除以其列数printf(%3d, k);printf(n);21、约瑟夫#include void main()int n, m, i, s=0; printf (Enter n: m: ); scanf(%d%d, &n, &m); for (i=2; i=n; i+) s=(s+m)%i; printf (The winner is %dn, s+1); 22、斐波拉契#includevoid main()long f, f1, f2; int i, n;printf(Enter n : ); scanf_s(%d, &n);f1 = 1; f2 = 1;printf(%10d%

31、10d, f1, f2);for (i = 1; i = n; i+)f = f1 + f2;printf(%10d, f);f1 = f2; f2 = f;if (i % 10 = 0)printf(n); 23、海滩上有一堆桃子，五只猴子来分。第一只猴子把这堆桃子凭据分为五份，多了一个，这只猴子把多的一个扔入海中，拿走了一份。第二只猴子把剩下的桃子又平均分成五份，又多了一个，它同样把多的一个扔入海中，拿走了一份，第三、第四、第五只猴子都是这样做的，问海滩上原来最少有多少个桃子？#includemain()int i, m, j, k, count;for (i = 4; i10000; i

32、 += 4)count = 0;m = i;for (k = 0; k5; k+)j = i / 4 * 5 + 1;i = j;if (j % 4 = 0)count+;else break;i = m;if (count = 4) printf(%dn, j);break;24、验证哥德巴赫猜想，即任一个偶数都可以分解为两个素数之和。#include int isprime(int);void even(int);void main()int a;printf(请输入一个偶数:); scanf_s(%d, &a);if (a % 2 = 0) even(a);else printf(%d

33、不是偶数！n);void even(int x)int i;for (i = 2; i = x / 2; i+)if (isprime(i) & isprime(x - i)printf(%d=%d+%dn, x, i, x - i); return;int isprime(int a)int i;for (i = 2; i = a / 2; i+)if (a%i = 0)return 0; return 1;25、魔方阵#include #define N 20void main()int aNN = 0 , i, j, k, n;doprintf(请输入魔幻方的阶数n(n%d):, N);

34、scanf_s(%d, &n);while (n = N | n % 2 = 0);i = n + 1; j = n / 2 + 1;a1j = 1; /将1放在第一行中间一列for (k = 2; k = n*n; k+)/*从2开始直到n*n各数依次按一下规则存放：每一个数存放的行比前一个数的行数减1，列数加1*/i-; j+;if (in)i += 2; j-;else if (i n) j = 1;/当上一个数的列数为n时，下一个数的列数应为1，行数减去1if (aij = 0)aij = k;/*如果按上面规则确定的位置上已有数，或上一个数是第一行第n列时，则把下一个数放在上一个数的下面。*/else i += 2; j-; aij = k; for (i = 1; i = n; i+)for (j = 1; j = n; j+)printf(%-4d, aij);printf(n);26、国际象棋棋盘#include #include void main() int i, j;SetConsoleOutputCP(437); /显示大于127的asc码for (i = 0; i 8; i+) for (j = 0; j 8; j+) if (i + j) % 2 = 0)printf(%c%c, 219, 219);elseprintf( );printf(n);