/*
第一题：
描述：
If we list all the natural numbers below 10
that are multiples of 3 or 5, we get 3, 5, 6 and 9.
The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
*/
#include <stdio.h>
#include <stdlib.h>
#define TEST 1 //编程之前就应该考虑测试(运行时将1改为0)
#if TEST
#define END_NUM 10 //问题给出的常数
#define SUM 23
#else
#define END_NUM 1000
#endif
#define BEGIN_NUM 1
#define FAC_1 3
#define FAC_2 5
int sum_of_mul (const int ,const int ,const int ,const int ) ;
int main( void )
{
printf ( "the sum of all the multiples" //太长的string literal
" of %d or %d below %d " , //可以用这种方法“分割”
FAC_1 , FAC_2 , END_NUM
);
printf ( "is %d.\n",
sum_of_mul ( BEGIN_NUM , END_NUM , FAC_1 , FAC_2 )
);
#if TEST
printf ( "= %d\n", SUM ) ;
#else
#endif
system("PAUSE");
return 0;
}
int sum_of_mul (const int begin ,const int end ,
const int fac1 ,const int fac2 )
{
int sum = 0 ;
int i ;
for( i = begin ; i < end ; i ++ )
{
if ( i % fac1 == 0 || i % fac2 == 0 )
{
sum += i ;
}
}
return sum ;
}
参考博文
http://www.cnblogs.com/zhouyinhui/archive/2011/01/05/1926769.html