小练习：数组元素的交换

/// 1.交换两个数组的元素，两个数组元素个数相同#define _CRT_SECURE_NO_WARNINGS 1#include<stdio.h>int main(){int arr1[] = { 2, 3, 5, 4, 6, 7, 9, 12, 34, 45 };int arr2[] = { 1, 9, 8, 0, 4, 6, 34, 56, 7, 23 };int i = 0, temp = 0;int size = sizeof(arr1) / sizeof(*arr1);//求数组元素的个数printf("交换前：\n");printf("arr1数组元素为：");for (i = 0; i < size; i++){printf("%4d", arr1[i]);}printf("\n");printf("arr2数组元素为：");for (i = 0; i < size; i++){printf("%4d", arr2[i]);}printf("\n");for (i = 0; i < size; i++){temp = arr1[i];arr1[i] = arr2[i];arr2[i] = temp;}printf("交换后：\n");printf("arr1数组元素为：");for (i = 0; i < size; i++){printf("%4d", arr1[i]);}printf("\n");printf("arr2数组元素为：");for (i = 0; i < size; i++){printf("%4d", arr2[i]);}fflush(stdin);getchar();return 0;}//2.求1/1 - 1/2 + 1/3 - 1/4......-1/100#define _CRT_SECURE_NO_WARNINGS 1#include<stdio.h>int main(){int i = 0;int flag = 1;double sum = 0;for (i = 1; i <= 100; i++){sum = sum + (1.0 / i)*flag;//(1.0/i)表示强制转换为浮点型flag = -flag;//实现加减交替//sum=sum+pow((-1),(1+i))*(1.0/i);//pow(a,b)为a的b次方，需要调用math.h}printf("求得结果为%lf", sum);getchar();return 0;}//3.求1到100里面9出现的次数#define _CRT_SECURE_NO_WARNINGS 1#include<stdio.h>int main(){int i = 0, sum = 0;for (i = 1; i <= 100; i++){if (i / 10 == 9)//十位上的9{sum++;}if (i % 10 == 9)//个位上的9{sum++;}}printf("1到100之间9总共出现了%d次", sum);getchar();return 0;}