BZOJ 2393: Cirno的完美算数教室 容斥原理

2393: Cirno的完美算数教室

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 557  Solved: 337
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Description

~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~

但是Cirno这么天才的妖精才不屑去数啦

只能依靠聪明的你咯。

Input

一行正整数L R

( 1 < L < R < 10^10)

Output

一个正整数，代表所求的答案

Sample Input

1 100

Sample Output

58

和1853一模一样啊

最后不能被其他baka数整除的baka数很少

用容斥原理，直接硬搞

#include<cmath>#include<ctime>#include<cstdio>#include<cstring>#include<cstdlib>#include<complex>#include<iostream>#include<algorithm>#include<iomanip>#include<vector>#include<string>#include<bitset>#include<queue>#include<map>#include<set>using namespace std;typedef long long ll;inline ll read(){ll x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch<='9'&&ch>='0'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}return x*f;}void print(ll x){if(x<0)putchar('-'),x=-x;if(x>=10)print(x/10);putchar(x%10+'0');}const int N=2010;ll l,r;int tot,n,ans;ll a[N];bool book[N];void getnum(ll x){if(x>r)return ;if(x)a[++tot]=x;getnum(x*10+2);getnum(10*x+9);}ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}void dfs(int x,int y,int z){if(!x){if(y&1)ans+=r/z-(l-1)/z;else if(y)ans-=r/z-(l-1)/z;return ;}dfs(x-1,y,z);ll t=z/gcd(z,a[x]);if((double)t*a[x]<=r)dfs(x-1,y+1,t*a[x]);}int main(){l=read();r=read();getnum(0);sort(a+1,a+1+tot);register int i,j;for(i=1;i<=tot;++i)if(!book[i]){a[++n]=a[i];for(j=1;j<=tot;++j)if(a[j]%a[i]==0)book[j]=1;}dfs(tot,0,1);cout<<ans<<endl;}/*1 10058*/