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;}