交换两个数组的内容+求1/1-1/2+1/3...-1/100的值+1-100整数中出现9的次数

交换两个数组的内容:

交换两个容量相同的整形数组的内容：

#include<stdio.h> /* 交换两个整形数组的内容 */
void print(int x[]) {
for (int i = 0; i < 3;i++) {
printf("%d “, x[i]);
}
printf(”\n");
}
int main() {
int a[3] = { 1, 2, 3 };
int b[3] = { 4, 5, 6 };
int c[3];
for (int i = 0; i < 3;i++) {
c[i] = a[i];
a[i] = b[i];
b[i] = c[i];
}
print(a);
print(b);
return 0;
}

交换两个字符串数组：

#include<stdio.h> /* 交换字符串 */
int main() {
char c1[6] = “YOU”;
char c2[6] = “SHE”;
char c3[6] = “”;
strcpy(c3, c1);
strcpy(c1, c2);
strcpy(c2, c3);
printf("%s\n", c1);
printf("%s\n", c2);
return 0;
}

求1/1-1/2+1/3…-1/100的值：

#include<stdio.h> /* 求1/1-1/2+1/3…-1/100的值 */
int main() {
int deno = 1; //分母
int coef = 1; //系数
double sum = 0.0;
//for (int i = 1; i <= 100;i++) {
// sum += coef * 1.0 / i;
// coef = -coef;
//}
while (deno <= 100) {
sum += coef * 1.0 / deno; //必须要用1.0浮点数，否则当deno>1时，1/deno作为整数计算为0
coef = -coef;
deno++;
}
printf(“The sum is %lf\n”, sum);
return 0;
}

统计1-100整数中出现9的次数：

#include<stdio.h> /* 1-100中有几个数字中含有9 */
int main() {
int count = 0;
for (int i = 1; i <= 100;i++) {
if (i == 9 || i%10 == 9) {
count++;
}
if (i / 10 == 9) {
count++;
}
}
printf("%d numbers have 9\n", count);
return 0;
}