Beautiful numbers

题目连接：https://vjudge.net/contest/365059#problem/A

题目大意 :
就是求区间内能被所有位上的数字(!0)整除的数的个数

想法：

首先lcm{1~9}=2520;

想到每个数都是1~9中某些数字的lcm 所以他们一定能整除2520

由数论知识可以知道： x % km % m = x % m

所以我们可以得到 x%2520%lcm{xi}==0  是满足条件

但是我们这个是以后发现如果根据这样开数组会导致空间不够！

每个数只能是1~9的最小公倍数 所以计算了下 所有的lcm一共有48种可 能 如下:

1 2 3 4 5 6 7 8 9 10 12 14 15 18 20 21 24 28 30 35 36 40 42 45 56 60 63 70 72 84 90 105 120 126 140 168 180 210 252 280 315 360 420 504 630 840 1260 2520

离散化一下就好了

#pragma GCC optimize(3,"Ofast","inline")//O3优化
#pragma GCC optimize(2)//O2优化
#include <algorithm>
#include <string>
#include <string.h>
#include <vector>
#include <map>
#include <stack>
#include <set>
#include <queue>
#include <math.h>
#include <cstdio>
#include <iomanip>
#include <time.h>
#include <bitset>
#include <cmath>
#include <sstream>
#include <iostream>
#include <cstring>

#define LL long long
#define ls nod<<1
#define rs (nod<<1)+1
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define INF 0x3f3f3f3f
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)

const double eps = 1e-10;
const int maxn = 2e5 + 10;
const int mod = 1e9 + 7;

int sgn(double a){return a < -eps ? -1 : a < eps ? 0 : 1;}
using namespace std;

LL L,R;
int sum,len;
int b[30];
LL mem[20][2522][50];
int k;
int vis[2550];

LL gcd(LL a,LL b) {
    LL t;
    while (b) {
        t = b;
        b = a % b;
        a = t;
    }
    return a;
}

LL  lcm(LL a,LL b) {
    return a  / gcd(a,b) * b;
}

void init() {
    k = 0;
    for (int i = 1;i <= 2520;i++) {
        if (2520 % i == 0)
            vis[i] = k++;
    }
}

LL dfs(int cur,int m,int Lcm,bool f) {
    if (cur < 0)
        return m % Lcm == 0;
    if (!f && mem[cur][m][vis[Lcm]] != -1)
        return mem[cur][m][vis[Lcm]];
    int v = 9;
    if (f)
        v = b[cur];
    LL ans = 0;
    for (int i = 0;i <= v;i++) {
        LL tlcm;
        if (i == 0)
            tlcm = Lcm;
        else
            tlcm = lcm(Lcm,i);
        ans += dfs(cur-1,(10*m+i)%2520,tlcm,f && (i == v));
    }
    if (!f)
        mem[cur][m][vis[Lcm]] = ans;
    return ans;
}

LL solve(LL x) {
    len = 0;
    while (x) {
        b[len++] = x % 10;
        x /= 10;
    }
    return dfs(len-1,0,1,1);
}

int main() {
    int T;
    scanf("%d",&T);
    init();
    memset(mem,-1,sizeof(mem));
    while (T--) {
        scanf("%I64d%I64d",&L,&R);
        LL ans = 0;
        ans = solve(R) - solve(L-1);
        printf("%I64d\n",ans);
//        cout << solve(R) << endl;
//        cout << solve(L-1) << endl;
    }
    return 0;
}

标签：Beautiful,GCC,include,2520,numbers,lcm,optimize,define
来源： https://www.cnblogs.com/-Ackerman/p/12615137.html