2013.12.26 流程控制相关练习奇偶数

最新推荐文章于 2022-05-02 00:43:24 发布

原创最新推荐文章于 2022-05-02 00:43:24 发布 · 526 阅读

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这篇博客探讨了如何计算1到100之间的所有奇数之和，是流程控制和数值计算的一个简单练习。

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求1到100之间的奇数和

# include <stdio.h>  
  
int main(void)  
{  
    int i;  
    int sum = 0;  
    for (i=1;i<=100;i+=2)  
    {  
        sum = sum + i;  
    }  
    printf("%d\n",sum);  
    return 0;  
}

求1到100奇数之和及奇数个数

# include <stdio.h>  
int main(void)  
{  
    int i;  
    int sum = 0;  
    int num = 0;  
    for (i=1; i<=100; i++)  
    {  
        if(i%2 != 0)  
        {  
            sum = sum + i;  
            num = num + 1;  
        }  
    }  
    printf("sum = %d\n",sum);  
    printf("num = %d\n",num);  
    return 0;  
}

1到100之间的奇数平均值

# include <stdio.h>  
int main(void)  
{  
    int i;  
    int sum = 0;  
    int num = 0;  
    int avarge;  
    for (i=1; i<=100; i++)  
    {  
        if(i%2 != 0)  
        {  
            sum = sum + i;  
            num = num + 1;  
        }  
    }  
    avarge = sum/num;  
    printf("avarge = %d\n",avarge);  
    return 0;  
}

求1到100之间的奇数和偶数之和

# include <stdio.h>
int main(void)
{
	int i;
	int sum1 = 0;//求奇数之和
	int sum2 = 0;//求偶数之和
	int sum = 0;
	for (i=1;i<=100;++i)
	{
		if(i%2 == 0)
		{
			sum2 = sum2 + i;
		}
		else
		{
			sum1 = sum1 + i;
		}
	sum = sum1 + sum2;
	}
	printf("sum =%d\n",sum);
	return 0;
}

求1+1/2+1/3+...+1/100的和

# include <stdio.h>  
  
int main(void)  
{  
    int i;  
    float sum = 0;//赋初值。  
    for (i=1;i<=100;++i)  
    {  
        sum = sum + 1/(float)i;//强制转换为浮点型，因为i定义成整型  
    }  
    printf("sum = %f\n",sum);//以%f输出  
    return 0;  
}

用while实现从1+2+3+...+100的和(for循环和while循环之间可以相互转化)

# include <stdio.h>
int main(void)
{
	int i;
	int sum = 0;
	i = 1;
	while(i<101)
	{
		sum = sum + i;
		i++;
	}
	printf("sum = %d\n",sum);
	return 0;
}