另类打表：将从1到n的满足美素数条件的数目赋值给prime[n]，这样最后仅仅须要用prime[L]减去prime[R-1]就可以；

美素数

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

Total Submission(s): 2332    Accepted Submission(s): 811

3

1 100

2 2

3 19

Case #1: 14

Case #2: 1

Case #3: 4

代码：

#include<stdio.h>

#include<string.h>

#define MAXN 1000005

int pri[MAXN] = {1,1};

int prime[MAXN];

int mei( int n )

{

	int i, sum = 0, pre = n;

	while(n){

		sum += n%10;

		n/=10;

	}

	if( pri[sum] == 0 )

	return 1;

	else

	return 0;

}

void f()

{

	for( int i = 2, k = 0; i < MAXN; i ++ ){

		if( pri[i] == 0 ){

			if( mei(i) )

			++k;

			for( int j = i+i; j < MAXN; j += i )

			pri[j] = 1;

		}

		prime[i] = k;

	}

}

int main()

{

	int l, r, t, ans, v = 1;

	scanf("%d", &t );

	f();

	while( t -- ){

		scanf( "%d%d", &l, &r );

		printf( "Case #%d: ", v++ );

		printf( "%d\n", prime[r]-prime[l-1] );

	}

	return 0;

}