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;
}