交換兩個陣列的內容:

交換兩個容量相同的整形陣列的內容：

#include<stdio.h>
/*
	交換兩個整形陣列的內容
*/
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;

}
 

交換兩個字串陣列：

#include<stdio.h>
/*
	交換字串
*/
int main() {

	//char c1[6];
	//strcpy(c1 , "LOVE");
	char c1[6] = "LOVE";
	char c2[6] = "SOCK";
	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;//係數
	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;//必須要用1.0浮點數，否則當deno>1時，1/deno作為整數計算為0
		coef = -coef;
		deno++;
	}
	printf("The sum is %lf\n", sum);
	return 0;

}
 

統計1-100整數中出現9的次數：

#include<stdio.h>
/*
	1-100中有幾個數字中含有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;

}