Pace 3 （1：交换两个数组内容，数组长度一致；2：计算1/1-1/2+1/3-...+1/999-1/100；3：计算0~100中9出现的次数）

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//交换两个数组的内容（数组长度一致）#define _CRT_SECURE_NO_WARNINGS 1#include <stdio.h>void Output(int arr[5],int len){for (int i = 0; i < len; i++){printf("%d ", arr[i]);}}int main(){int a[5] = { 1, 2, 3, 4, 5 };int b[5] = { 6, 7, 8, 9, 0 };int t , i;int s = sizeof(a)/sizeof(a[0]);printf("数组的长度为%d\n",s);printf("交换前的数组为：\na数组：");Output(a,s);printf("\nb数组：");Output(b,s);for (i = 0; i < s; i++){t = a[i];a[i] = b[i];b[i] = t;}printf("\n交换后的数组为：\na数组：");Output(a,s);printf("\nb数组：");Output(b,s);system("pause");return 0;}

//计算1/1-1/2+1/3-...+1/999-1/100#define _CRT_SECURE_NO_WARNINGS 1#include <stdio.h>int main(){int i;float sum = 0, a = 0; for (i = 1; i <= 100; i++){if (i % 2 == 0){a = -i;}else a = i;sum = sum+1.0/a;}printf("%f ", sum);system("pause");return 0;}

//计算0~100中9出现的次数#define _CRT_SECURE_NO_WARNINGS 1#include <stdio.h>int main(){int i ,sum = 0;int a = 0, b = 0;for (i = 1; i <= 100; i++){a = i / 10;//取出十位if (a == 9) sum++;b = i - a * 10;//取出个位if (b == 9) sum++;}printf("%d\n", sum);system("pause");return 0;}

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