c语法问题，在n值下打印3和5的倍数之和[关闭]

Question

为什么会出现语法错误，我认为没有任何问题。谢谢.. 语法错误在第 2 行。

#include <stdio.h>
int main() {
    long int sum;
    int an;
    printf("Sum of multiples of three and five\nEnter the nth value\n>> ");
    scanf("%d",&an);
    sum=((3/2*an/3*(an/3+1))+(5/2*an/5*(an/5+1))-(15/2*an/15*(an/15+1)));
    printf("The sum of multiples of 3 and 5 under %d is %ld",an,sum);
}

Answer 1

缺少的; 是一个很容易修复的印刷错误，但这不是发布代码中的主要问题。

sum = ((3/2*an/3*(an/3+1))+(5/2*an/5*(an/5+1))-(15/2*an/15*(an/15+1)));
//      ^^^   ^     ^       ^^^   ^     ^       ^^^^   ^    ^^^^^   

此行无法提供正确答案，因为多个整数除法会“截断”（对整数类型执行操作，实际上不涉及浮点变量）中间值到整数。

通过将此评估拆分为几个函数，您可能会在代码可读性方面有所收获。

long int sum_of_multiples(int factor, int n)
{
    // Number of multiples of factor up to n.
    // E.g. 13 / 3 = 4  --> 3, 6, 9, 12 
    long int multiples = n / factor;
    
    // Sum of the multiples using Gauss's method.
    // 3 + 6 + 9 + 12 = 3 * (1 + 2 + 3 + 4) = 3 * (4 * (4 + 1)) / 2
    // Note that the product of an odd and an even number is always even
    return factor * (multiples * (multiples + 1) / 2);
}

long int sum_of_multiples_of_3_and_5_less_than_n(int n)
{
    if ( n < 1 )
        return 0;
    // The OP wrote "under", so I assume they want to exclude the number itself
    --n;
    return sum_of_multiples(3, n) + sum_of_multiples(5, n) - sum_of_multiples(15, n);
}

Answer 2

将 ; 放在第 7 行后，它工作正常。