函数fun()的功能是:统计所有小于等于n的素数的个数,素数的个数作为函数值返回.这段程序是怎么达到目的#include int fun(int n){int i,j,count=0;printf("\nThe prime number between 3 to %d\n",n);for (i=3; i=i){count+
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![函数fun()的功能是:统计所有小于等于n的素数的个数,素数的个数作为函数值返回.这段程序是怎么达到目的#include int fun(int n){int i,j,count=0;printf(](/uploads/image/z/3692466-18-6.jpg?t=%E5%87%BD%E6%95%B0fun%28%29%E7%9A%84%E5%8A%9F%E8%83%BD%E6%98%AF%EF%BC%9A%E7%BB%9F%E8%AE%A1%E6%89%80%E6%9C%89%E5%B0%8F%E4%BA%8E%E7%AD%89%E4%BA%8En%E7%9A%84%E7%B4%A0%E6%95%B0%E7%9A%84%E4%B8%AA%E6%95%B0%2C%E7%B4%A0%E6%95%B0%E7%9A%84%E4%B8%AA%E6%95%B0%E4%BD%9C%E4%B8%BA%E5%87%BD%E6%95%B0%E5%80%BC%E8%BF%94%E5%9B%9E.%E8%BF%99%E6%AE%B5%E7%A8%8B%E5%BA%8F%E6%98%AF%E6%80%8E%E4%B9%88%E8%BE%BE%E5%88%B0%E7%9B%AE%E7%9A%84%23include+int+fun%28int+n%29%7Bint+i%2Cj%2Ccount%3D0%3Bprintf%28%22%5CnThe+prime+number+between+3+to+%25d%5Cn%22%2Cn%29%3Bfor+%28i%3D3%3B+i%3Di%29%7Bcount%2B)
函数fun()的功能是:统计所有小于等于n的素数的个数,素数的个数作为函数值返回.这段程序是怎么达到目的#include int fun(int n){int i,j,count=0;printf("\nThe prime number between 3 to %d\n",n);for (i=3; i=i){count+
函数fun()的功能是:统计所有小于等于n的素数的个数,素数的个数作为函数值返回.这段程序是怎么达到目的
#include
int fun(int n)
{int i,j,count=0;
printf("\nThe prime number between 3 to %d\n",n);
for (i=3; i=i)
{count++; printf(count%15?"%5d":"\n%5d",i);}
}
return count;
}
main()
{int n=20,r;
r = fun(n);
printf("\nThe number of prime is :%d\n",r);
}
#include
int fun(int n)
{int i,j,count=0;
printf("\nThe prime number between 3 to %d\n",n);
for (i=3; i=i) 这里j>=i是起什么作用?
{count++; printf(count%15?"%5d":"\n%5d",i);}
}
return count;
}
main()
{int n=20,r;
r = fun(n);
printf("\nThe number of prime is :%d\n",r);
}
函数fun()的功能是:统计所有小于等于n的素数的个数,素数的个数作为函数值返回.这段程序是怎么达到目的#include int fun(int n){int i,j,count=0;printf("\nThe prime number between 3 to %d\n",n);for (i=3; i=i){count+
if (i%j == 0)
break; 这句用break是什么意思?
如果i被j整除.即i除以j而没有余数,
说明不是质数,而是合数.所以用break跳出 for (j=2; j=i) 这里j>=i是起什么作用?
说明从2到j都没有能够整除i,所以是质数啊,就加一了.
你理清这里的算法思想.
是以外层循环来取得某数,在以内层循环来判断此数是否是质数.
是的话内层循环能够执行完毕,从而计数加1
不是的话,就会跳出内层循环,继续在外层循环取数