题目链接：
codeforces 55D
题目大意：
求在[l,r]中能够整除自己每个数位上的数字的数的个数。
题目分析：AC代码：;LL;LL dp[20][50][3000];int d[20],hash[3000];const int mod = 2520;int gcd ( int a ,int b ){return !b?a:gcd(b,a%b);}void init ( ){memset ( dp , -1 , sizeof ( dp ) );int cc = 0;for ( int i = 1 ; i <= 8 ; i *= 2 )for ( int j = 1; j <= 9 ; j*= 3 )for ( int k = 1 ; k <= 5 ; k *= 5 )for ( int t = 1; t <= 7 ; t *= 7 )hash[i*j*k*t] = ++cc;}LL dfs ( int n , int lcm , int f , int r ){if ( !n ) return r%lcm==0;int ll = hash[lcm];if ( f && dp[n][ll][r] != -1 )return dp[n][ll][r];int x = f?9:d[n];LL ret = 0;for ( int i = 0 ; i <= x ;i++ )ret += dfs ( n-1 , i==0?lcm:lcm*i/gcd(lcm,i) , i==x?f:1 , (r*10+i)%mod );return f? dp[n][ll][r] = ret : ret;}LL solve ( LL x ){int cc = 0;while ( x ){d[++cc] = x%10;x /= 10;}return dfs ( cc , 1 , 0 , 0 );}int main ( ){int t;LL a,b;scanf ( “%d” , &t );init ( );while ( t– ){scanf ( “%lld%lld” , &a , &b );printf ( “%lld\n” , solve ( b ) – solve ( a-1 ) );}}
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