【数位DP】 codeforces 55D && FZU chriswho

sign。。。很难的数位DP。。。。。

#include <iostream>
#include <queue> 
#include <stack> 
#include <map> 
#include <set> 
#include <bitset> 
#include <cstdio> 
#include <algorithm> 
#include <cstring> 
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 1005
#define maxm 200005
#define eps 1e-10
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid 
#define rson o<<1 | 1, mid+1, R
//#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}
LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}
// head

const int tt = 2520;
LL dp[20][3000][50];
int x[50], mpp[3000];
int s[50];
int cnt;

void init(void)
{
	cnt = 0;
	for(int i = 1; i <= tt; i++)
		if(tt % i == 0) x[++cnt] = i, mpp[i] = cnt;
	memset(dp, -1, sizeof dp);
}

LL dfs(int pos, int sum, int lcm, int limit)
{
	if(pos == 0) {
		if(sum % x[lcm] == 0) return 1;
		else return 0;
	}
	if(dp[pos][sum][lcm] != -1 && !limit) return dp[pos][sum][lcm];
	lcm = x[lcm];
	int b = (limit ? s[pos] : 9);
       	LL ans = 0;
	for(int i = 0; i <= b; i++) {
		int nlcm;
		if(!i) nlcm = lcm;
		else nlcm = lcm / gcd(lcm, i) * i;
		ans += dfs(pos - 1, (sum * 10 + i) % tt, mpp[nlcm], limit && i == b);
	}
	if(!limit) dp[pos][sum][mpp[lcm]] = ans;
	return ans;
}

LL calc(LL t)
{
	int size = 0;
	while(t) {
		s[++size] = t % 10;
		t /= 10;
	}
	return dfs(size, 0, 1, 1);
}

int main(void)
{
	int _;
	LL a, b;
	init();
	while(scanf("%d", &_)!=EOF) {
		while(_--) {
			scanf("%I64d%I64d", &a, &b);
			printf("%I64d\n", calc(b) - calc(a - 1LL));
		}
	}

	return 0;
}