ZOJ 3436 July Number(DFS)-蒲公英云

ZOJ 3436 July Number(DFS)

题意把一个数替换为这个数相邻数字差组成的数知道这个数仅仅剩一位数若最后的一位数是7 则称原来的数为 July Number 给你一个区间求这个区间中July Number的个数

从7開始DFS 位数多的数总能由位数小的数推出

#include <bits/stdc++.h>
using namespace std;
const int N = 1e6;
int july[N], n;
set<int> ans;
int pw[] = {1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000};
//剩余长度, 当前位要填的数，通过什么数来搜索, 路径
void dfs(int len, int d, int bas, int cur)
{
    if(d < 0 || d > 9) return;  //要填的数不合法
    if(!len)
    {
        july[n++] = cur * 10 + d;
        return;
    }
    int k = pw[len - 1];
    dfs(len - 1, d - bas / k, bas % k, cur * 10 + d);
    dfs(len - 1, d + bas / k, bas % k, cur * 10 + d);
}
int main()
{
    ans.insert(7);
    set<int>::iterator it;
    for(int l = 2; l < 10; ++l)
    {
        n = 0;
        for(it = ans.begin(); it != ans.end(); ++it)
            for(int i = 1; i < 10; ++i)
                dfs(l - 1, i, *it, 0);
        for(int i = 0; i < n; ++i) ans.insert(july[i]);
        //printf("%d\n", ans.size());
    }
    int a, b = n = 0;
    for(it = ans.begin(); it != ans.end(); ++it) 
        july[n++] = *it;
    while(~scanf("%d%d", &a, &b))
        printf("%d\n", upper_bound(july, july + n, b) - lower_bound(july, july + n, a));
    return 0;
}

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].