素数判定
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 166313 Accepted Submission(s): 58802
Problem Description
对于表达式n^2+n+41,当n在(x,y)范围内取整数值时(包括x,y)(-39<=x<y<=50),判定该表达式的值是否都为素数。
Input
输入数据有多组,每组占一行,由两个整数x,y组成,当x=0,y=0时,表示输入结束,该行不做处理。
Output
对于每个给定范围内的取值,如果表达式的值都为素数,则输出"OK",否则请输出“Sorry”,每组输出占一行。
Sample Input
0 1 0 0
Sample Output
OK
Author
lcy
Source
Recommend
答案:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int judge(int sum)
{
int mid = (int)sqrt((double)sum);//sqrt函数为开方函数,参数必须为double
int i;
for (i=2;i<=mid;i++)
{
if(sum%i==0)
return 1;
}
return 0;
}
void main()
{
int n,m;
int sum;
int i;
int sign;
while (~scanf("%d %d",&n,&m))
{
if (n==0&&m==0)
break;
sign = 0;
for(i=n;i<=m;i++)
{
sum = i*i+i+41;
sign = judge(sum);
if(sign == 1)
break;
}
if(sign==1)
printf("Sorry\n");
else
printf("OK\n");
}
}