經(jīng)典C語言源代碼

1、C 語言源代碼1、 (1)某年某月某日是星期幾#includeint main()int year, month, day;while (scanf_s(%d%d%d, &year, &month, &day) != EOF)if (month = 1 | month = 2)d.%d, &start0, &start1, &start2);printf( 請輸入結(jié)束日期，如：n);scanf_s(%d.%d.%d, &end0, &end1, &end2);int sum = 0;for (int mid = start0; mid end0; mid+) if (mid % 400 = 0

2、) | (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之間有 d 天n, start0,start1,start2,end0,end1,end2, sum);int indexday(int year, int month, int day) int i, flag, dayth;int month_day = 0,31,28

3、,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、遞歸求1*1+2*2+3*3+n*n#include long Element(int n) if (n = 1)return 1 * 1;elsereturn Element(n - 1) + n*n;voi

4、d main() int n;printf( 請輸入 n 的值： n);scanf_s(%d, &n);printf( 所求值為%dn, Element(n);19、最大公約數(shù)(輾轉(zhuǎn)相除)#includevoid main() /* 輾轉(zhuǎn)相除法求最大公約數(shù)*/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) /* 余數(shù)不為0，繼續(xù)相除，直到余數(shù)為0 */c = a%b; a = b; b = c;printf(The largest

5、 common divisor:%dn, a);printf(The least common multiple:%dn, m*n / a);20、楊輝三角#includevoid 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;/ 每次要打印的下一個數(shù)等于前一個數(shù)乘以其所在行數(shù)和列數(shù)的差再除以其列數(shù)printf(%3d, k);printf(n);21、約瑟夫#includ

6、e 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%10d, f1, f2);for (i = 1; i = n; i+)f = f1 + f2;printf

7、(%10d, f);f1 = f2; f2 = f;if (i % 10 = 0)printf(n);23、海灘上有一堆桃子，五只猴子來分。第一只猴子把這堆桃子憑據(jù)分為五份，多了一個，這只猴子把多的一個扔入海中，拿走了一份。第二只猴子把剩下的桃子又平均分成五份，又多了一個，它同樣把多的一個扔入海中，拿走了一份，第三、第四、第五只猴子都是這樣做的，問海灘上原來最少有多少個桃子？#includemain()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

8、 + 1;i = j;if (j % 4 = 0) count+;else break;i = m;if (count = 4) printf(%dn, j);break;24、 驗證哥德巴赫猜想，即任一個偶數(shù)都可以分解為兩個素數(shù)之和。#include int isprime(int);void even(int);void main()int a;printf( 請輸入一個偶數(shù):);scanf_s(%d, &a);if (a % 2 = 0)even(a);elseprintf(%d 不是偶數(shù)！n);void even(int x)int i;for (i = 2; i = x / 2; i

9、+)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、魔方陣#include #define N 20void main()int aNN = 0 , i, j, k, n;doprintf( 請輸入魔幻方的階數(shù)n(n%d):, N);scanf_s(%d, &n);while (n = N | n % 2 = 0);i = n + 1;j

10、= n / 2 + 1;a1j = 1; / 將 1 放在第一行中間一列for (k = 2; k = n*n; k+)/*從2開始直到n*n各數(shù)依次按一下規(guī)則存放：每一個數(shù)存放的行比前一個數(shù)的行數(shù)減1，列數(shù)加1*/i-; j+;if (in)i += 2; j-;else if (i n) j = 1;/ 當(dāng)上一個數(shù)的列數(shù)為n 時，下一個數(shù)的列數(shù)應(yīng)為1，行數(shù)減去 1if (aij = 0)aij = k;/* 如果按上面規(guī)則確定的位置上已有數(shù)， 或上一個數(shù)是第一行第n 列時， 則把下一個數(shù)放在上一個數(shù)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、國際象棋棋盤#include #include void ma