- 将数组A中的内容和数组B中的内容进行交换。(数组一样大)
#include<stdio.h>
#include<stdlib.h>
int main()
{
int a[6] = { 1, 2, 3, 4, 9, 8 };
int b[6] = { 2, 5, 6, 5, 3, 4 };
int i=0;
while (i < 6)
{
int tmpe;
tmpe = a[i];
a[i] = b[i];
b[i] = tmpe;
i++;
}
i = 0;
printf("a[i]=");
while (i <= 5)
{
printf("%d ",a[i]);
printf("\n");
printf("b[i]=");
i = 0;
while (i <= 5)
{
printf("%d ", b[i]);
i = i + 1;
}
printf("\n");
system("pause");
}
- 计算1/1-1/2+1/3-1/4+1/5 …… + 1/99 - 1/100 的值。
#include<stdio.h>
#include<stdlib.h>
int main()
{
double sum1 = 0, sum2 = 0, sum = 0, num1 = 1, num2 = 2;
for (num1 = 1; num1 <= 99; num1 += 2)
{
sum1 = sum1 + 1 / num1;//奇数分之一之和
}
for (num2 = 2; num2 <= 100; num2 += 2)
{
sum2 = sum2 + 1 / num2;//偶数分之一之和
}
sum = sum1-sum2;//总和为sum1减去sum2
printf("sum=%f/n", sum);//输出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=0,times=0;//定义变量num和times
for(num=0;num<=100;num=+1)
{
if (num%10==9)//个位数有9次数
{
times=times+1;
}
if(num/10=9)//十位有9的次数
{
times=times+1;
}
}
printf(“times=%d",times);
system("pause");
}
提交作业