ZOJ3436 July Number 【DFS】

July Number

---

Time Limit: 2 Seconds      Memory Limit: 65536 KB

---

The digital difference of a positive number is constituted by the difference between each two neighboring digits (with the leading zeros omitted). For example the digital difference of 1135 is 022 = 22. The repeated digital difference, or differential root, can be obtained by caculating the digital difference until a single-digit number is reached. A number whose differential root is 7 is also called July Number. Your job is to tell how many July Numbers are there lying in the given interval [a,b].

Input

There are multiple cases. Each case contains two integers a and b. 1 ≤a ≤b ≤ 10⁹.

Output

One integer k, the number of July Numbers.

Sample Input

1 10

Sample Output

1

题意:(1)题意：一个数，前后两个数之差组成新数，一直下去，如果结果是7，则为7月数。如7 18 29 70 81 92 108 118 188 198 209 219 229 299 700770 780 790 801 811 881 891 902 910 912 922 980 992
这些都是7月数。给一个区间，问这个区间中有几个7月数。

    题目不难，但代码实现有些麻烦。

#include<cstring>#include<iostream>#include<cstdio>#include<algorithm>#include<vector>using namespace std;vector<int> ans,v;int p[11]={0,1,10,100,1000,10000,100000,1000000,10000000,100000000};void dfs(int len,int i,int x,int cut){    cut=cut*10+i;    if(len==0)    {        v.push_back(cut);        return ;    }    int tem=x/p[len];    x=x-tem*p[len];    if(i+tem<=9)dfs(len-1,i+tem,x,cut);    if(i-tem>=0)dfs(len-1,i-tem,x,cut);}void slove(){    ans.push_back(7);    for(int len=1;len<9;len++)    {        v.clear();        for(int i=1;i<=9;i++)        {            for(int j=0;j<ans.size();j++)            {//cout<<"　　***";                dfs(len,i,ans[j],0);            }        }        ans.insert(ans.end(),v.begin(),v.end());        sort(ans.begin(),ans.end());        ans.erase(unique(ans.begin(),ans.end()),ans.end());    }}int main(){    slove();    int x,y;    while(scanf("%d%d",&x,&y)!=EOF)    {        printf("%d\n",upper_bound(ans.begin(),ans.end(),y)-lower_bound(ans.begin(),ans.end(),x));    }    return 0;}