BZOJ 1853 幸运数字(容斥原理+dfs)

题意:求闭区间内能被6和8组成的数字整除的数目。n<=1e11.

我们可以预处理出这些6和8组成的数字，大概2500个，然后排除一些如88，66的情况。这样大概还剩下1000个。

转化为[0,r]和[0,l-1]的问题，显然需要运用容斥原理。ans=n/6+n/8+n/68+...+...-n/lcm(6,8)-n/lcm(6,68)......

因此用dfs即可计算出来，这样一看复杂度好像是2^1000的样子，但是注意到lcm增长的很快，如果lcm>n那么显然之后的这些情况就可以忽略了。

这就是一个强有力的剪枝。

另外从大到小dfs要比从小到大dfs要好。大概常数小？

 

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-9
# define MOD 1024523
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=500005;
//Code begin...

LL num[2505], pos1, pos2, p[15], mark[1005];
void init(){
    p[0]=1; FOR(i,1,10) p[i]=p[i-1]*10;
    int l=1, r=2, tmpl, tmpr;
    num[++pos1]=6; num[++pos1]=8;
    FOR(i,2,10) {
        tmpl=r+1;
        FOR(j,l,r) num[++pos1]=6*p[i-1]+num[j];
        FOR(j,l,r) num[++pos1]=8*p[i-1]+num[j];
        tmpr=pos1;
        l=tmpl; r=tmpr;
    }
    FOR(i,1,pos1) {
        int flag=true;
        FO(j,1,i) if (num[i]%num[j]==0) {flag=false; break;}
        if (flag) mark[++pos2]=num[i];
    }
    mark[++pos2]=1e16;
}
LL dfs(int pos, int flag, LL x, LL cheng){
    if (pos<=0) return 0;
    LL res=0;
    res+=dfs(pos-1,flag,x,cheng);
    LL tmp=__gcd(cheng,mark[pos]);
    if (cheng/tmp<=(double)x/mark[pos]) {
        LL tt=cheng/tmp*mark[pos];
        res+=dfs(pos-1,flag^1,x,tt);
        res+=(flag?x/tt:-x/tt);
    }
    return res;
}
LL sol(LL x){
    for (int i=pos2; i>=1; --i) if (mark[i]<=x) return dfs(i,1,x,1);
    return 0;
}
int main ()
{
    init();
    LL a, b;
    scanf("%lld%lld",&a,&b);
    printf("%lld\n",sol(b)-sol(a-1));
    return 0;
}