C program 程式問題－dlxpxv7的部落格

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C program 程式問題

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1.The LCM(Lowest Common Multiple) of two numbers x, y is the smallest whole number which is divisible by x and y. The following algorithm istrivial though it is not the most efficient.Step 1: Store x*y to mStep 2:If m is divisible by both x and y, store m to LCMStep 3:Decrease m by 1Step 4:Repeat... 顯示更多 1. The LCM(Lowest Common Multiple) of two numbers x, y is the smallest whole number which is divisible by x and y. The following algorithm is trivial though it is not the most efficient. Step 1: Store x*y to m Step 2:If m is divisible by both x and y, store m to LCM Step 3:Decrease m by 1 Step 4:Repeat steps 2 and 3 as long as m is greater than both x and y. Step 5:Output LCM Write a C program to input two positive numbers and find their LCM. The program should give the following sample output: Enter two numbers: 12, 18 The LCM of 12 and 18 is 36 2.Write a program that find the sum of the digits of a number entered through the keyboards. Your program should cater for both positive and negative numbers. examples: Enter a number : 2511 sum of digits = 9 Enter a number : -2511 sum of digits = 9 更新: 001 - 答非所問,你講緊嘢呀? 002 - sorry, 個program 真係run唔到,可唔可以打多次呀,thanks!! 更新 2: 第2题俾少少貼士呀? 主要問题係点樣將輸入x位數字加晒?謝謝! 更新 3: sorry, really don't understand = =

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Here is the code for question 1. Feel free to PM me if you have questions. I will leave Q2 for you as exercise, make use of guide given by the previous post. Good luck. /* http://hk.knowledge.yahoo.com/question/question?qid=7008112401110 ianianian02 2008-11-24 17:36:39 1. The LCM(Lowest Common Multiple) of two numbers x, y is the smallest whole number which is divisible by x and y. The following algorithm is trivial though it is not the most efficient. Step 1: Store x*y to m Step 2:If m is divisible by both x and y, store m to LCM Step 3:Decrease m by 1 Step 4:Repeat steps 2 and 3 as long as m is greater than both x and y. Step 5:Output LCM Write a C program to input two positive numbers and find their LCM. The program should give the following sample output: Enter two numbers: 12, 18 The LCM of 12 and 18 is 36 */ //#include #include #include int main() { // This algorithm is definitely NOT very efficient, but here is the code // In addition, the algorithm is WRONG, because step 4 should read: // Step 4:Repeat steps 2 and 3 as long as m is NOT LESS THAN x or y. // The correction is required when x is a multiple of y, such as 48, 24 long n1=-1, n2=-1, product,LCM; while(n1n1&&product>n2) LCM of 27 & 54 would give 108 and not 54 while(product>=n1&&product>=n2) { if(product%n1==0&&product%n2==0)LCM=product; product--; } printf("The LCM of %ld and %ld is %d ",n1,n2,LCM); system("PAUSE"); return 0; } 2008-11-26 08:28:34 補充： What compiler do you use, what is the error message the computer gives? I am sorry that the additional answers would not give enough space to show the program again. For the second problem, 001 has given a few useful hints. Here's some more: 2008-11-26 08:28:41 補充： 1. if the number is negative, convert it to positive. Define a variable call sum and set it to zero. 2. use the % (modulus) function to find the last digit. Add it to sum. 3. Divide it by 10 to remove the last digit. 4. repeat 2 and 3 until the number is zero. PM me if you need more information.

其他解答:

1. Step2至4 以for廻圈進行: for (m = x * y; m > x && m > y; m--) Step2的檢查條件: if ((m % x == 0) && (m % y == 0)) 2. 以字串/char陣列儲存用家輸入以strlen找出用家輸入的字串的長度如假設輸入的數字並沒有空白, 先測試一下字串中零位元素是否"-" 如零位元素是"-", 則輸出sum of digits為strlen結果減一否則, 直接輸出strlen作為sum of digits