数组交换、计算求值、出现9的次数

1.将数组A中的内容和数组B中的内容进行交换。（数组一样大） 
# include<stdio.h>
int main()
{
	int i = 0;
	int j = 0;
	int a[3] = { 3, 5, 7 };
	int b[3] = { 9, 8, 2 };
	for (i = 0; i < 3; i++)
	{
		a[i] = a[i] ^ b[i];
		b[i] = a[i] ^ b[i];
		a[i] = b[i] ^ a[i];
	}
	for (j = 0; j < 3; j++)
	{
		printf("a[%d]=%d\n", j, a[j]);
	}
	for (j = 0; j < 3; j++)
	{
		printf("b[%d]=%d\n", j, b[j]);
	}

    return 0;
}
2. 计算1/1-1/2+1/3-1/4+1/5 …… + 1/99 - 1/100 的值
# include<stdio.h>
int main()
{
	double a = 0;
	double b = 0;
	double c = 0;
	double sum = 0;
	double   i = 0;
	double   j = 0;
	for (i = 1; i <= 99; i += 2)
	{
		a  = a+(1 / i);
	}
	for (j = 2; j <= 100; j += 2)
	{
		b =b+ ((-1) / j);
	}
		sum = a + b;
	printf("sum=%lf", sum);
	return 0;
}
3. 编写程序数一下 1到 100 的所有整数中出现多少次数字9
# include<stdio.h>
int main()
{
	int i = 0;
	int j = 0;
	int k = 0;
	int count = 0;
	for (i = 1; i <= 100; i++)
	{
		j = i % 10;
		k = i / 10;
		if (j == 9)
			count++;
		if (k == 9)
		count++;
	}
	printf("出现9的次数为%d\n", count);
	return 0;
}