#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<stdlib.h>
int main()
{
int arr1[3] = { 1, 2, 3 };
int arr2[3] = { 4, 5, 6 };
int count = sizeof(arr1) / sizeof(0);
int i;
for (i = 0; i < count; i++)
{
arr1[i] = arr1[i] + arr2[i];
arr2[i] = arr1[i] - arr2[i];
arr1[i] = arr1[i] - arr2[i];
}
printf("arr1[3] = ");
for (i = 0; i < count; i++)
{
printf("%d ", arr1[i]);
}
printf("\n");
printf("arr2[3] = ");
for (i = 0; i < count; i++)
{
printf("%d ", arr2[i]);
}
printf("\n");
system("pause");
}
计算1/1-1/2+1/3-1/4+1/5 …… + 1/99 - 1/100 的值。
#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<stdlib.h>
#include<math.h> //直接调用库函数即可
int main()
{
int i;
double sum = 0.0;
for (i = 1; i <= 100; i++)
{
sum = sum + (pow(-1, i + 1)) / i; //利用pow函数控制系数
}
printf("sum = %f\n", sum);
system("pause");
}
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>
#include<stdlib.h>
int main()
{
int num;
int count = 0;
for (num = 1; num <= 100; num++)
{
if (num % 10 == 9)
{
printf("%d ", num);
count = count + 1;
}
if (num / 10 == 9)
{
printf("%d ", num);
count = count + 1;
}
}
printf("\n");
printf("count = %d", count);
printf("\n");
system("pause");
}