经历是一种收获1200字:C语言考题,速!

来源:百度文库 编辑:杭州交通信息网 时间:2024/05/03 05:16:35
求一维整形数组中负数的个数
要求:
a[10]={-15,-22,13,44,-25,66,0,88-39,10}求负数的个数

#include <stdio.h>
void main()
{
int a[10]={-15,-22,13,44,-25,66,0,88-39,10};
int i,n;
for (i=0,n=0;i<10;i++) if (a[i]<0) n++;
printf("负数个数=%d\n",n);
}

#include <stdio.h>

int main()
{
int a[10]={-15, -22, 13, 44, -25, 66, 0, 88, -39, 10};
int n = 0;
for (int i = 0; i < 10; i++)
{
if (a[i] < 0)
{
n++;
}
}
printf("%d个负数", n);
return 0;
}

int iCount = 0;
int i = 0;
for (i = 0; i < 10; i++)
{
if (a[i] < 0)
{
iCount++;
}
}

iCount中存的就是负数的个数

#include <stdio.h>

int main()
{
int a[10]={-15, -22, 13, 44, -25, 66, 0, 88, -39, 10};
int n = 0;
for (int i = 0; i < 10; i++)
{
if (a[i] < 0)
{
n++;
}
}
printf("%d个负数", n);
return 0;
}

#include<stdio.h>
int main(void)
{

int a[10]={-15,-22,13,44,-25,66,0,88-39,10},count=0,minus;
for (;count<10;((a[count++]<0)?minus++:minus))
;

printf(\"total minus:%d\",minus);
}