经典C语言源代码VIP

1、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 2 year-;month += 12;int c = year / 100;int y = year - c * 100;int week = (c / 4) - 2 * c + (y + y / 4) + (13 * (month + 1) / 5) day - 1;whil

2、e (week0) week += 7; week %= 7;switch (week)case 1:printf(Mondayn); break;case 2:pri ntf(Tuesdayn); break;case 3:pri ntf(Wed nesdayn); break;case 4:pri ntf(Thursdayn); break;case 5:printf(Fridayn); break;case 6:pri ntf(Saturdayn); break;case 0:pri ntf(S un day n); break;return 0;1、（2）某年某月某日是第几天（一维数组

3、）#in clude stdio.h void mai n() int i, flag, year, mon th, 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);sca nf_s(%d/%d/%d, & year, &mon th, &day);dayth = day;flag = (year % 400 = 0) | (year % 4 = 0 & year % 100 != 0);if (flag)mon th_day2 = 29;for (i = 1; i mon

4、 th; i+)dayth = dayth + mon th_dayi;prin tf(%d/%d/%d 是第 %d 天 n, year, mon th, day, dayth);2、30个数中找最小的数及其位置#i nclude stdio.h# define SIZE 30void mai n() int i;float dataSIZE;int min;printf(请输入%d个浮点数：n,SIZE);for (i = 0; i SIZE; i+) /scan f_s(%f, & datai);datai = ran d() % 30 + 1;printf(%f、, datai);min

5、 = 0;for (i = 1; i SIZE; i+) if (datai data min)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)#i nclude stdio.h# define SIZE 30void mai n() 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+) sca nf_s(%f, & datai);for (i = 0; i SIZE; i+) min = i;for

6、(j = i + 1; j SIZE; j+) if (dataj datami n)min = j;temp = datami n;datami n = 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)#i nclude stdio.h# define SIZE 5void accept_array(float a, i nt size);void sort(float a, i nt size);void show_array(float

7、 a, i nt size);void mai n() 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, i nt size) int i;printf(请输入 %d 个分数：, size);for (i = 0; i size; i+)sca nf_s(%f, & ai);void show_array(float a,

8、i nt size) int i;for (i = 0; i size; i+)printf( %5.2f, ai);prin tf(n);void sort(float a,i nt size) int i, mi n, j;float temp;for (i = 0; i SIZE; i+) min = i;for (j = i + 1; j SIZE; j+)if (aj amin)欢迎下载5min = j;temp = ami n;ami n = ai;ai = temp;4、( 1)指针加减：#i nclude stdio.h#define SIZE 10void mai n() i

9、nt aSIZE = 1,2,3,4,5,6,7,8,9,10 ;int *pa, i;pa = & a0;pa=a;prin tf(n);for (i = 0; i SIZE; i+) printf( %d, *pa);printf(%d, *(pa+1);pa+;(2)指针比较：#i nclude stdio.h#define SIZE 10void mai n() 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;

10、 pa+)sca nf_s(%d, pa);for (pa-; qa = pa; pa-)printf( %d, *pa);5、两字符串相连：#i nclude stdio.h#i nclude stri ng.hvoid str_cat(char str1, char str2);void mai n() 欢迎下载#int i, j；char str1160;char str280;printf(请输入第一个字符串：);gets(strl);printf(请输入第二个字符串：);gets(str2);str_cat(str1, str2);puts(strl);void str_cat(ch

11、ar str1, char str2) int i, j;i = 0;while (str1i != 0)i+;j = 0;while (str2j != 0) str1i = str2j;i+; j+;str1i = 0：6、二维数组(a, b转置)#i nclude stdio.hvoid mai n() 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;prin tf(na:n);for (i = 0; i 3; i+) for (j = 0; j 2; j+)pr

12、in tf(%5d, aij);prin tf(n);prin tf(nb:n);for(i = 0; i 2; i+) for (j = 0; j 3; j+)prin tf(%5d, bij);prin tf(n);欢迎下载#7、输入一个二维数组并输出(指针)#i nclude stdio.hvoid mai n() int x23;int i, j；for (i = 0; i 2; i+)for (j = 0; j 3; j+)sca nf_s(%d, *(x + i) + j); putchar(n);for (i = 0; i 2; i+)for (j = 0; j 3; j+)pr

13、in tf(%d, *(*(x + i) + j);putchar(n);8冒泡法排序一个数组#in clude stdio.h#defi ne size 10void maopao(i nt a);void mai n() int a10;int i;printf(请输入10个整数：n);for (i = 0; i 10; i+)sca nf_s(%d, & ai);maopao(a);void maopao(i nt 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

14、+ 1;/大的沉底，小的上浮 aj + 1 = temp;printf(排序结果：n);for (i = 0; i 10; i+)printf(%4d, ai);9、两数组A，B,要求AB,如A：4,79B：1，3,5,8,9变换后A：1,3,5B：4，78,9,9#in elude 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)欢迎下载13x =

15、 Am - 1;Am - 1 = B0;/ 交换 Am-1和 B0j = 1；while (jn & Bj= 0 & Aix)Ai + 1 = Ai-;/寻找B0的插入位置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+) sea nf_s(%d, &Ai);for (i = 0; i5; i+) sea nf_s(%d, & Bi);ReArra nger(A, B, 3, 5);printf(n输出第一个数组：);for (i = 0; i3; i+)pri

16、n tf(%d , Ai);prin tf(nn);printf(输出第二个数组：);for (i = 0; i5; i+)prin tf(%d , Bi);prin tf(n);10、符合 1+6+3=3+2+5=1+4+5 有哪几组A1B C64D E F 到325#in clude void mai n() int a, b, c, d, e, f;for (a = 1; a = 6; a+)for (b = 1; b = 6; b+) if (b = a)con ti nue;for (c = 1; c = 6; c+) if (c = a) | (c = b)con ti nue;f

17、or (d = 1; d = 6; d+) if (d = a) | (d = b) | (d = c)con ti nue;for (e = 1; e = 6; e+) if (e = a) | (e = b) | (e = c) | (e = d)con ti nue;f = 21 - (a + b + c + d + e);if (a + b + d = d + e + f) & (a + b + d = a + c + f) prin tf( %dn, a);prin tf(%d%dn, b, c);11、输入一串字符，升序排序，折半查找其中一字符#in elude void *sor

18、tStri ng(char un sort, i nt len gth) for (i nt i = 0; i len gth; i+) for (i nt j = i + 1; j un sortj) int temp = un sortj;un sortj = un sorti;un sorti = temp;void mai n() 欢迎下载17char s150;gets(sl);char value;sca nf_s(%c, &value);printf(s1 : %sn, s1);prin tf(value:%cn, value);int len gth = strle n(s1)

19、;prin tf(le ngth:%dn, le ngth);sortStri ng(s1, le ngth);printf(s1 : %sn, s1);int start = 0, end = len gth - 1;int mid = (end + start) / 2;while (start s1mid)start = mid+1;elseend = mid-1;if (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直接任整除#in cludeint isprime(i nt n)if (n 2)return 0

20、;for (i nt i = 2; in / 2; i+)if (n%i = 0)return 0;return 1;void mai n()int i, k = 0;for (i = 100; i = 300; i+)if (isprime(i)prin tf(%3d , i);k+;if (k % 10 = 0) prin tf(n);for (i = 500; i = 700; i+)if (isprime(i)prin tf(%3d , i);k+;if (k % 10 = 0) prin tf(n);prin tf(n);13、判断一个数是否是素数#in clude#in clude

21、void mai n()int m, i, k;printf(请输入一个整数:);sea nf_s(%d, & m);k = (in t)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#in clude #in clude bool is5Num(i nt num) int temp = num % 10;while (temp != 5 & num 10) num = num / 10;temp = num % 10;欢迎下载21if (temp != 5)return fa

22、lse;elsereturn true;void mai n() int num;printf(输入一个数：n);sca nf_s(%d, &n um);if (is5Nu m(num)printf(含 5n);elseprintf(不含 5n);15、一个排好序的数组，插入一个数#in clude void mai n() int a11 = 1,2,3,4,5,6,7,8,9,10 ;int num;printf(插入前数组为：n);for (i nt i = 0; i a9)a10 = num;else for (int i = 0; i 10;i+)if (num = i; j-)aj

23、 + 1 = aj;ai = num; break;printf(插入后数组为：n);for (i nt i = 0; i = 10; i+)prin tf(%3d, ai);prin tf(n);16、牛顿迭代法：#in elude #in elude double fune(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 preeision) 迭代次数double x1;

24、int k;if (fun e1(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 - fun e(x0) / fun c1(x0); 进行牛顿迭代计算while (!(f un e1(x1 - x0) preeisi on | fabs(fu nc(x1) preeisi on) x0 = x1;/准备下一次迭代欢迎下载23if (fun c1(xO) = 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 - fun c(x0) / fun c1(x0);/ 进行牛顿迭代计算 v

25、oid mai n() double x, precisi on;printf(输入初始迭代值x0： n);sca nf_s(%lf, & x);printf(迭代要求的精度：n);sca nf_s(%lf, &precisi on);Newton(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#i nclude stdio.hvoid mai n() int start3, end3;printf(请输入开始日期，如1964219: n);seanf_s(%d.%d.%d,

26、&startO, &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 (i nt mid = start0; mid en d0; mid+) if (mid % 400 = 0) | (mid % 4 = 0 & mid % 100 != 0) sum = sum + 366;elsesum = sum + 365;sum = sum - in dexday(start0,start1,start2) +in dexday(e nd0,e

27、 nd1,e nd2);printf(在%d.%d.%d-%d.%d.%d之间有 %d 天n,start0,start1,start2,end0,end1,end2, sum);int in dexday(i nt year, i nt mon th, i nt day) 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);欢迎下载25if (flag)mo

28、 nth_day2 = 29;for (i = 1; i mon th; i+)dayth = dayth + mon th_dayi; return dayth;18、递归求 1*1+2*2+3*3+n*n#i nclude stdio.hlong Eleme nt(i nt n) if (n = 1)return 1 * 1;elsereturn Eleme nt(n - 1) + n*n;void mai n() int n;printf(请输入n的值：n);sca nf_s(%d, &n);19、最大公约数(辗转相除)#in cludevoid mai n()/*辗转相除法求最大公约数

29、*/int m, n, a, b, t, c;prin tf(I nput two in teger nu mbers: n);sca nf_s(%d%d, &a, & b);m = a; n = b;0 */while (b != 0)/*余数不为0,继续相除，直到余数为c = a%b; a = b; b = c;prin tf(The largest com mon divisor:%dn, a);prin tf(The least com mon multiple:%dn, m*n / a);20、杨辉三角#in cludevoid mai n() int i, j, n, k;prin

30、 tf(E nter n:); sea nf_s(%d, &n);for (i = 1; i = n; i+)k = 1;for (j = 1; ji; j+)prin tf(%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 和列数的差再除以其列数prin tf(%3d, k);prin tf(n);21、约瑟夫#i nclude void mai n()int n, m, i, s=0;printf (E nter n: m:);sea nf(%d%d, &n, & m);for (i=2; i=n; i+) s=(s+m)%i;prin

31、tf (The winner is %dn, s+1); 22、斐波拉契#in cludevoid mai n()long f, f1, f2; int i, n;printf(Enter n :);sca nf_s(%d, &n);f1 = 1;f2 = 1;prin tf(%10d%10d, f1, f2);for (i = 1; i = n; i+)f = f1 + f2;prin tf(%10d, f);f1 = f2; f2 = f;if (i % 10 = 0) prin tf(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

32、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？#in cludemai n()int i, m, j, k, count;for (i = 4; i10000; i += 4)count = 0;m = i;for (k = 0; k5; k+)j = i / 4 * 5 + 1;i = j;if (j % 4 = 0) coun t+;else break;i = m;if (co unt = 4) prin tf(%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