poj 3909 --Lucky numbers(搜索)

 

 

题源：http://poj.org/problem?id=3909

 

 

题意：very lucky number 定义：我们称只包含 4和 7的为lucky number ,

由一个或多个 lucky number的乘积得到的数我们就叫做very lucky number  

 

解法：只由4和7组成的 只有8000多个

 

先找出这些数 (lucky number )

 

再由这些数搜索出 very lucky number  

 

大概有35W个 

 

 

//3520K125MS#include <iostream>#include <algorithm>using namespace std;__int64 num[1000000];int cnt=0;__int64 Maxn=1000000000000;__int64 a,b,sum;char str[20];void dfs(int n){if(n==12)return ;str[n]='4';str[n+1]=0;sscanf(str,"%I64d",&num[cnt++]);dfs(n+1);str[n]='7';str[n+1]=0;sscanf(str,"%I64d",&num[cnt++]);dfs(n+1);}void dfs2(__int64 t,int pos){for(int i=pos;i<sum;i++){__int64 temp=t*num[i];if(temp>0&&temp<=Maxn){num[cnt++]=temp;dfs2(temp,i);}elsebreak;}}void init(){//search the number only contain '4' and '7'dfs(0);sum=cnt;sort(num,num+cnt);//srarch the all lucky number include 10^12dfs2(1,0);sort(num,num+cnt);//delete repeated numbercnt=unique(num,num+cnt)-num;num[cnt]=10000000000010;// printf("%d/n",cnt);// for(int i=0;i<11;i++)// printf("%I64d/n",num[i]);}//besrarch //the fist number larger or equal than a//the fist number larger than bint Tsearch(__int64 t,int index){int low=0,high=cnt,mid,i;while (low+1<high){mid=(low+high)>>1; if(num[mid]<=t)low=mid;elsehigh=mid;}if(index==1)//B{for(int i=low;;i++)if(num[i]>t)return i;}else //A{for( i=low;;i++)if(num[i]>=t)return i;}}int solve(){int ans;ans=Tsearch(b,1)-Tsearch(a,0);if(ans<0)ans=0;return ans;}int main(){ init(); int t; scanf("%d",&t); while(t--) { scanf("%I64d%I64d",&a,&b); printf("%d/n",solve()); }}