BZOJ_2393_Cirno的完美算数教室&&BZOJ_1853_[Scoi2010]幸运数字 _深搜+容斥原理
题意：
~Cirno发现了一种baka数，这种数呢~只含有2和⑨两种数字~~
现在Cirno想知道~一个区间中~~有多少个数能被baka数整除
分析：
基本思路是找到1e10以内所有的baka数，如果a[i]%a[j]==0则把a[i]删掉
每个方案的lcm的贡献是r/lcm-(l-1)/lcm，深搜每一种情况，用容斥原理确定正负
考虑优化，从大到小枚举baka数，当lcm大于r时返回
代码：
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
#define LL long long
LL a[2000]={0,6ll,8ll,68ll,86ll,668ll,686ll,866ll,868ll,886ll,6668ll,6686ll,6866ll,6886ll,8666ll,8668ll,8866ll,66668ll,66686ll,66866ll,66868ll,66886ll,68666ll,68668ll,68686ll,68866ll,86666ll,86668ll,86686ll,86866ll,88868ll,88886ll,666668ll,666686ll,666866ll,666868ll,666886ll,668666ll,668686ll,668866ll,686666ll,686668ll,686866ll,688666ll,688868ll,688886ll,866666ll,866668ll,866686ll,868666ll,868886ll,886666ll,886868ll,888668ll,888686ll,888866ll,888868ll,888886ll,6666668ll,6666686ll,6666866ll,6666868ll,6668666ll,6668668ll,6668686ll,6668866ll,6686666ll,6686668ll,6686686ll,6686866ll,6688666ll,6688868ll,6688886ll,6866666ll,6866668ll,6866686ll,6866866ll,6868666ll,6868868ll,6868886ll,6886666ll,6886868ll,6886886ll,6888668ll,6888866ll,6888868ll,6888886ll,8666666ll,8666686ll,8666866ll,8668666ll,8668868ll,8686666ll,8686868ll,8686886ll,8688668ll,8688686ll,8688866ll,8688868ll,8688886ll,8866666ll,8866868ll,8866886ll,8868668ll,8868686ll,8868866ll,8868868ll,8868886ll,8886668ll,8886686ll,8886866ll,8886868ll,8886886ll,8888666ll,8888668ll,8888686ll,8888866ll,66666668ll,66666686ll,66666866ll,66666868ll,66666886ll,66668666ll,66668668ll,66668686ll,66668866ll,66686666ll,66686686ll,66686866ll,66688666ll,66688868ll,66688886ll,66866666ll,66866668ll,66866866ll,66868666ll,66868868ll,66886666ll,66886868ll,66886886ll,66888668ll,66888686ll,66888866ll,66888868ll,66888886ll,68666666ll,68666668ll,68666686ll,68668666ll,68668868ll,68668886ll,68686666ll,68686886ll,68688668ll,68688686ll,68688866ll,68688868ll,68688886ll,68866666ll,68866868ll,68868668ll,68868686ll,68868866ll,68868868ll,68886668ll,68886686ll,68886866ll,68886868ll,68886886ll,68888668ll,68888686ll,68888866ll,86666666ll,86666668ll,86666686ll,86666866ll,86668868ll,86668886ll,86686666ll,86686868ll,86686886ll,86688686ll,86688866ll,86688868ll,86688886ll,86866666ll,86866868ll,86866886ll,86868668ll,86868868ll,86868886ll,86886668ll,86886686ll,86886866ll,86886868ll,86886886ll,86888666ll,86888668ll,86888686ll,86888866ll,88666666ll,88666868ll,88666886ll,88668686ll,88668868ll,88668886ll,88686668ll,88686686ll,88686866ll,88686868ll,88686886ll,88688666ll,88688668ll,88688686ll,88688866ll,88866668ll,88866866ll,88866868ll,88866886ll,88868666ll,88868668ll,88868686ll,88868866ll,88886666ll,88886668ll,88886686ll,88886866ll,88888666ll,88888868ll,88888886ll,666666686ll,666666866ll,666666868ll,666666886ll,666668666ll,666668686ll,666668866ll,666686666ll,666686668ll,666686686ll,666686866ll,666688666ll,666688868ll,666688886ll,666866666ll,666866668ll,666866686ll,666866866ll,666868666ll,666868868ll,666868886ll,666886666ll,666886868ll,666886886ll,666888668ll,666888686ll,666888866ll,666888868ll,666888886ll,668666666ll,668666668ll,668666686ll,668666866ll,668668666ll,668668868ll,668668886ll,668686666ll,668686868ll,668686886ll,668688668ll,668688686ll,668688866ll,668688868ll,668688886ll,668866666ll,668866868ll,668866886ll,668868668ll,668868686ll,668868868ll,668868886ll,668886668ll,668886686ll,668886866ll,668886868ll,668886886ll,668888666ll,668888668ll,668888686ll,668888866ll,686666666ll,686666668ll,686666686ll,686666866ll,686668666ll,686668868ll,686668886ll,686686666ll,686686868ll,686686886ll,686688668ll,686688686ll,686688866ll,686688868ll,686688886ll,686866666ll,686866868ll,686866886ll,686868668ll,686868686ll,686868866ll,686868868ll,686868886ll,686886668ll,686886686ll,686886866ll,686886868ll,686886886ll,686888666ll,686888668ll,686888686ll,686888866ll,688666868ll,688666886ll,688668668ll,688668686ll,688668866ll,688668868ll,688668886ll,688686668ll,688686686ll,688686866ll,688686868ll,688686886ll,688688666ll,688688668ll,688688866ll,688866668ll,688866686ll,688866866ll,688866868ll,688866886ll,688868666ll,688868668ll,688868866ll,688886666ll,688886668ll,688886686ll,688886866ll,688888666ll,688888868ll,688888886ll,866666666ll,866666668ll,866666686ll,866666866ll,866668666ll,866668868ll,866668886ll,866686666ll,866686868ll,866686886ll,866688668ll,866688686ll,866688866ll,866688868ll,866688886ll,866866666ll,866866868ll,866866886ll,866868668ll,866868686ll,866868866ll,866868868ll,866868886ll,866886668ll,866886686ll,866886866ll,866886868ll,866886886ll,866888666ll,866888668ll,866888866ll,868666666ll,868666868ll,868666886ll,868668668ll,868668686ll,868668866ll,868668868ll,868686668ll,868686686ll,868686866ll,868686868ll,868686886ll,868688668ll,868688686ll,868688866ll,868866866ll,868866868ll,868866886ll,868868666ll,868868668ll,868868686ll,868868866ll,868886666ll,868886668ll,868886686ll,868886866ll,868888666ll,868888868ll,868888886ll,886666666ll,886666868ll,886666886ll,886668668ll,886668866ll,886668868ll,886668886ll,886686668ll,886686686ll,886686866ll,886686868ll,886686886ll,886688666ll,886688668ll,886688686ll,886688866ll,886866668ll,886866686ll,886866866ll,886866868ll,886866886ll,886868666ll,886868668ll,886868686ll,886868866ll,886886666ll,886886686ll,886886866ll,886888666ll,886888868ll,886888886ll,888666668ll,888666686ll,888666866ll,888666868ll,888666886ll,888668666ll,888668668ll,888668686ll,888668866ll,888686666ll,888686668ll,888686686ll,88868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int cnt=943;
LL L,R,ans;
LL gcd(LL n,LL m){
return m?gcd(m,n%m):n;
}
void dfs(int dep,LL val,int tot){
if(dep==1){
if(tot==0)return ;
tot&1?ans+=(R/val-L/val):ans-=(R/val-L/val);return ;
}
LL tmp=val/gcd(a[dep-1],val)*a[dep-1];
dfs(dep-1,val,tot);
if(tmp<0||tmp>R)return ;
dfs(dep-1,tmp,tot+1);
}
int main(){
scanf("%lld%lld",&L,&R);
L--;
int pos=0;
dfs(cnt+1,1,0);
printf("%lld
",ans);
}
（代码是1853，数据较强）