[UVa]10235:Simply Emirp

這題就照題意，
建質數表，
檢查輸入值是否為質數，
再檢查倒轉過來是否跟原本的值是否不相同且還是質數。

P.S. 這題我因為題意沒看清楚爆了Orz
我沒仔細看到翻轉過來要另外一個質數，而不能跟原來一樣Orz...
可愛的different......

[C](0.036)

	#include<stdio.h>
	int prime[1000005] = { 1, 1, 0 };
	int main()
	{
	int i, j;
	for( i = 2 ; i <= 1000000 ; i++ )
	if( !prime[i] )
	for( j = i+i ; j <= 1000000 ; j+=i )
	prime[j] = 1;
	int N;
	while( scanf( "%d", &N ) != EOF )
	{
	int N_origin = N;
	int N_reverse = 0;
	while( N )
	{
	N_reverse *= 10;
	N_reverse += N % 10;
	N /= 10;
	}
	if( prime[N_origin] )
	printf( "%d is not prime.\n", N_origin );
	else if( prime[N_reverse] || N_origin == N_reverse )
	printf( "%d is prime.\n", N_origin );
	else
	printf( "%d is emirp.\n", N_origin );
	}
	return 0;
	}

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