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来源:互联网 发布:matlab 7 for linux 编辑:程序博客网 时间:2024/05/01 18:35
将数组A和数组B内容交换:
#include <stdio.h>
void swap(int*p1, int*p2)
{
int tmp = *p1;
*p1 = *p2;
*p2 = tmp;
}
int main()
{
int arr1[10] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
int arr2[10] = { 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 };
int a = 0;
printf("arr1=");
for (a = 0; a < 10; a++)
printf(" %d ",arr1[a]);
printf("\n");
int b = 0;
printf("arr2=");
for (b = 0; b < 10; b++)
printf(" %d ", arr2[b]);
int i = 0;
for (i = 0; i < 10; i++)
swap(&arr1[i], &arr2[i]);
printf("\n");
int c = 0;
printf("arr1=");
for (c = 0; c < 10; c++)
printf(" %d ", arr1[c]);
printf("\n");
int d = 0;
printf("arr2=");
for (d = 0; d < 10; d++)
printf(" %d ", arr2[d]);
printf("\n");
return 0;
}
计算1/1-1/2+1/3-1/4+1/5 …… + 1/99 - 1/100 的值:
#include <stdio.h>
int main()
{
int i = 0;
int num = 0;
num = 1;
for (i = 2; i <= 100; i+=2)
num = num - 1 / i;
int j = 0;
for (j = 3; j <= 100; j += 2)
num = num + 1 / j;
printf("1 / 1 - 1 / 2 + 1 / 3 - 1 / 4 + 1 / 5 …… + 1 / 99 - 1 / 100=%d\n", num);
return 0;
}
int main()
{
int i = 0;
int num = 0;
num = 1;
for (i = 2; i <= 100; i+=2)
num = num - 1 / i;
int j = 0;
for (j = 3; j <= 100; j += 2)
num = num + 1 / j;
printf("1 / 1 - 1 / 2 + 1 / 3 - 1 / 4 + 1 / 5 …… + 1 / 99 - 1 / 100=%d\n", num);
return 0;
}
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>
int main()
{
int coun = 0;
int tmp = 0;
for (tmp = 1; tmp <= 99; tmp++)
{
if (tmp / 10 == 9)
coun++;
if (tmp % 10 == 9)
coun++;
}
printf("共有 %d 个9\n", coun);
return 0;
}
{
int coun = 0;
int tmp = 0;
for (tmp = 1; tmp <= 99; tmp++)
{
if (tmp / 10 == 9)
coun++;
if (tmp % 10 == 9)
coun++;
}
printf("共有 %d 个9\n", coun);
return 0;
}