2 seconds
256 megabytes
standard input
standard output
Robbers, who attacked the Gerda's cab, are very successful in covering from the kingdom police. To make the goal of catching them even harder, they use their own watches.
First, as they know that kingdom police is bad at math, robbers use the positional numeral system with base 7. Second, they divide one day in n hours, and each hour in m minutes. Personal watches of each robber are divided in two parts: first of them has the smallest possible number of places that is necessary to display any integer from 0 to n - 1, while the second has the smallest possible number of places that is necessary to display any integer from 0 to m - 1. Finally, if some value of hours or minutes can be displayed using less number of places in base 7 than this watches have, the required number of zeroes is added at the beginning of notation.
Note that to display number 0 section of the watches is required to have at least one place.
Little robber wants to know the number of moments of time (particular values of hours and minutes), such that all digits displayed on the watches are distinct. Help her calculate this number.
The first line of the input contains two integers, given in the decimal notation, n and m (1 ≤ n, m ≤ 109) — the number of hours in one day and the number of minutes in one hour, respectively.
Print one integer in decimal notation — the number of different pairs of hour and minute, such that all digits displayed on the watches are distinct.
2 3
4
8 2
5 链接：http://codeforces.com/contest/686/problem/C晚上做的时候想到了位数大于7的时候，输出0，但就是没想到之后暴力就可以解了 = =
还有，这代码中的判断条件想法也比较好，开一个used的vector，之后判断每个数变成7进制之后的各个数出现的次数，如果其中最大的小于等于1，那么就可行。这种想法真的很好，写着也简洁，需要学习。
#include <bits/stdc++.h>
using namespace std; int main()
{
//这是加速cin的，网上说用完之后，速度和scanf差不多
iostream::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr); //size_t 相当于无符号int
size_t n,m;
cin>>n>>m;
size_t len1=,len2=;
for (size_t a=;a<n;a*=)
len1++;
for (size_t b=;b<m;b*=)
len2++; size_t ans=;
if (len1+len2<=)
for (size_t i=;i<n;i++)
for (size_t j=;j<m;j++)
{
vector<size_t> used(,);
for (size_t a=i,k=;k<len1;a/=,k++)
{
used[a%]+=;
}
for (size_t b=j,k=;k<len2;k++,b/=)
{
used[b%]+=;
} //max_element 返回迭代器，所以要解引用
if (*max_element(used.begin(),used.end())<=)
ans++;
}
cout<<ans<<endl; return ;
}