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].
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
Author: HE, Ningxu
Contest: ZOJ Monthly, November 2010
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