L.C.M
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2023-12-01
LCM或最小公共两个值的多个,是两个值的倍数的最小正值。
For example ,3和4的倍数是 -
3 →3,6,9,12,15 ......
4 →4,8,12,16,20 ......
两者的最小倍数为12,因此3和4的最小公倍数为12。
算法 (Algorithm)
该程序的算法可以推导为 -
START
Step 1 → Initialize A and B with positive integers
Step 2 → Store maximum of A & B to max
Step 3 → Check if max is divisible by A and B
Step 4 → If divisible, Display max as LCM
Step 5 → If not divisible then step increase max, goto step 3
STOP
伪代码 (Pseudocode)
现在让我们为这个程序派生伪代码 -
procedure even_odd()
Initialize A and B
max = max(A, B)
WHILE TRUE
IF max is divisible by A and B THEN
LCM = max
BREAK
ENDIF
Increment max
END WHILE
DISPLAY LCM
end procedure
实现 (Implementation)
该算法的实现如下 -
#include<stdio.h>
int main() {
int a, b, max, step, lcm;
a = 3;
b = 4;
lcm = 0;
if(a > b)
max = step = a;
else
max = step = b;
while(1) {
if(max%a == 0 && max%b == 0) {
lcm = max;
break;
}
max += step;
}
printf("LCM is %d", lcm);
return 0;
}
输出 (Output)
该方案的产出应该是 -
LCM is 12