一直在超时，无奈交表。有时候交表是一个不错的办法~
以下是我的代码：
#include<stdio.h>
int main()
{
long a[8][100]={
{2,3,5,7},
{23,29,31,37,53,59,71,73,79},
{233,239,293,311,313,317,373,379,593,599,719,733,739,797},
{2333,2339,2393,2399,2939,3119,3137,3733,3739,3793,3797,5939,7193,7331,7333,7393},
{23333,23339,23399,23993,29399,31193,31379,37337,37339,37397,59393,59399,71933,73331,73939},
{233993,239933,293999,373379,373393,593933,593993,719333,739391,739393,739397,739399},
{2339933,2399333,2939999,3733799,5939333,7393913,7393931,7393933},
{23399339,29399999,37337999,59393339,73939133}
};
int n,i;
scanf("%d",&n);
i=0;
while( a[n-1][i]!=0 )
{
printf("%ld\n",a[n-1][i]);
i++;
}
return 0;
}
posted on 2010-01-06 19:17
lee1r 阅读(249)
评论(0) 编辑 收藏 引用 所属分类:
题目分类：数学/数论