[Program 81] Title: 809 * ?? = 800 * ?? 9 * ?? 1 Among them, the two digits of the representative, 8 * ?? the result of two digits, 9 * ?? the result of 3 Bit number. Request for the two digits of the representative, and the result of 809 * ??. 1. Program analysis: 2. Source code: OUTPUT (long b, long i) {printf ("/ n% ld /% ld = 809 *% ld % ld", B, I, I, b% i); } main () {long Int A, B, I; A = 809; for (i = 10; i <100; i ) {b = i * a 1; if (b> = 1000 && b <= 10000 && 8 * i < 100 && 9 * i> = 100) Output (b, i);}} ================================== ============================ 【Program 82】 Title: Octa conversion to decimal 1. Program analysis: 2. Program source code: main () {char * p, s [6]; int N; p = S; Gets (p); n = 0; while (* (p)! = '/ 0') {n = n * 8 * P -'0 '; p ;} Printf ("% d", n);} =============================== ======================================= 【Program 83】 Title: Seeking the odd number of odd numbers that can be composed of 0-7.
1. Program analysis: 2. Source code: main () {long sum = 4, s = 4; int J; for (j = 2; j <= 8; j ) / * j is place of number * / { Printf ("/ n% ld", sum); if (j <= 2) s * = 7; else * = 8; SUM = S;} Printf ("/ nsum =% ld", sum);} == ============================================================================================================================================================================================================= ========== [Program 84] Title: A number of even numbers can be represented as the sum of two prime numbers.
1. Program analysis: 2. Source code: #include "stdio.h" #include "math.h" main () {Int A, B, C, D; scanf ("% d", & a); for B = 3; B <= a / 2; b = 2) {for (c = 2; c <= SQRT (B); C ) IF (b% c == 0) Break; if (c> sqrt (B )) D = ab; elsebreak; for (c = 2; c <= SQRT (D); C ) IF (D% c == 0) Break; if (c> SQRT (D)) Printf ("% D = % D % D / N ", A, B, D);}} =================================== ============================= 【Program 85】 Title: Judgment a number of numbers can be taken 9 整 1. Program analysis: 2 . Source code: main () {long int m9 = 9, sum = 9; int zi, n1 = 1, c9 = 1; scanf ("% d", & zi); while (n1! = 0) {IF ! (sum% zi)) N1 = 0; ELSE {m9 = m9 * 10; sum = SUM M9; C9 ;}} printf ("% ld, can be divided by% D /" 9 / ", SUM, C9);} ============================================== ================ 【Program 86】 Title: Two string connections 1. Program analysis: 2. Source code: #include "stdio.h" main () { Char a [] = "acegikm"; char b [] = "bdfhjlnpq"; char C [80], * p; int i = 0, j = 0, k = 0; while (a [i]! = '/ 0 '&& B [J]! =' / 0 ') {IF (a [i] {c [k] = a [i]; i ;} elsec [k] = b [j ]; k ;} C [K ] = '/ 0'; if (a [i] == '/ 0') P = B J; ELSEP = A I;
STRCAT (C, P); PUTS (c);} ======================================= ========================= 【Program 87】 Title: Answer results (Structure variable delivery) 1. Program analysis: 2. Program source code: #include "stdio.h" struct student {int x; char C;} a; main () {ax = 3; ac = 'a'; f (a); printf ("% D,% C", AX, AC);} f (struct student b) {bx = 20; bc = 'y';} ============================ =================================== 程序 程序 88] Topic: 7 digits (1-50) The integer value, each reads a value, the program prints the number of this value. 1. Program analysis: 2. Source code: main () {INT i, a, n = 1; while (n <= 7) {DO {scanf ("% d", & a);} while (a <1 || a> 50); for (i = 1; i <= a; i ) printf ("*"); printf ("/ n"); n ;} getCH ();} ======= ============================================================================================================================================================================================================= ===== [Program 89] Topic: A company uses public telephone delivery data, data is an integer, encrypted in the delivery process, encrypted rules are as follows: Each number plus 5, then use and Instead of this number, the first bit and the fourth interchange are exchanged, and the second bit and the third one are exchanged.
