ZOJ3436 July Number

July Number

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

題解：

題目大意：將一個數字的相鄰兩位的差（的絕對值）組成一個新的數字，不斷重複，如果最後得到7，就稱這個數爲July Number，比如9024 – 922 – 70 – 7。題目要求1e9範圍內給定區間[a, b]裏July Number的個數。

這種數字其實不是很多,通過運行結果得到以1-9結尾的個數

  1------451156127           2------131145647            3------44286220       4-----14800278     5------4397180

  6------1031969  7------160218         8-----13851    9-----256

因此直接先算出所有數存起來排序後，然後二分查找即可，！比如由7出發，可以得到70, 81, 92, 19, 29，從19出發可以得到109, 890，這隻要枚舉最後以爲數字，然後枚舉相鄰兩位的差是正的還是負的，可以簡單的寫成一個dfs。注意每次都是用已算出的所有數來逆推，而不是用上一輪的結果。

附上AC代碼

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <string>
#include <queue>
#define inf 0x3f3f3f3f
typedef long long ll;
using namespace std;
const int maxn = 1005;

vector<ll> a;
vector<ll> v;
ll di[] = {10,100,1000,10000,100000,1000000,10000000,100000000,1000000000};
int getarr(ll a,int b[])<span style="white-space:pre">	</span>//數字轉化爲數組
{
    int t = 0;
    while(a != 0)
    {
        b[t++] = a%10;
        a /= 10;
    }
    return t-1;
}
ll getnum(int a[],int n)<span style="white-space:pre">	</span>//數組轉化爲數字
{
    ll ans = 0, d = 1;
    for(int i = 0;i < n;++i)
    {
        ans += a[i]*d;
        d *= 10;
    }
    return ans;
}

void dfs(int l,int r,int a[],int tp[])
{
    if(l > r-2)
    {
        ll ans = getnum(tp,r);

        if(ans >= di[r-2]){
            v.push_back(ans);
        }
        return;
    }
    int k1 = tp[l]+a[l];
    if(0 <= k1 && k1 < 10)
    {
        tp[l+1] = k1;
        dfs(l+1,r,a,tp);
    }
    int k2 = tp[l]-a[l];
    if(0 <= k2 && k2 < 10 && k1 != k2)
    {
        tp[l+1] = k2;
        dfs(l+1,r,a,tp);
    }
}

void fun(int hi,int s)
{
    int a[20] = {0}, tp[20] ={0};
    getarr(s,a);
    for(int i = 0;i < 10;++i)<span style="white-space:pre">	</span>//枚舉最後一位來推前面的每一位的數
    {
        tp[0] = i;
        dfs(0,hi,a,tp);
    }
}
void solve()
{
    a.push_back(7);
    for(int i = 2;i < 10;++i)<span style="white-space:pre">		</span>//枚舉位數
    {
        v.clear();
        for(int j = 0;j < a.size();++j)<span style="white-space:pre">	</span>//枚舉已經得到的數
        {
            fun(i,a[j]);
        }
        a.insert(a.end(),v.begin(),v.end());
        sort(a.begin(),a.end());
        a.erase(unique(a.begin(),a.end()),a.end());
    }
}

int main()
{
    //freopen("1.txt","w",stdout);
    solve();
    int x, y;

    /*
    for(int i = 0;i < a.size();++i)
        cout<<a[i]<<endl;
        */
    while (scanf("%d%d", &x, &y) != EOF) {
		printf("%d\n", upper_bound(a.begin(), a.end(), y) - lower_bound(a.begin(), a.end(), x));
	}
    return 0;
}