等式（a + b）<= n中的'a'的打印值可被x整除
给定方程式，程序必须找到'a'的值，其中a+b<=n并且也可以被x整除。
算法
START
Step 1 -> Declare start variables b=10, x=9, n=40 and flag=0, divisible
Step 2 -> Loop For divisible = (b / x + 1 ) * x and divisible <= n and divisible += x
IF divisible - b >= 1
Print divisible-1
Set flag=1
End
END
STOP示例
#include <stdio.h>
int main(int argc, char const *argv[]) {
int b=10, x=9, n=40, flag = 0;
int divisible;
for (divisible = (b / x + 1 ) * x ; divisible <= n; divisible += x) {
if ( divisible - b >= 1) {
printf("%d ", divisible - b );
flag = 1;
}
}
return 0;
}输出结果
如果我们运行上面的程序，那么它将生成以下输出
8 17 26