Codeforces Round #359 (Div. 2) C. Robbers' watch (暴力DFS)

题目链接：http://codeforces.com/problemset/problem/686/C

给你n和m，问你有多少对(a, b) 满足0<=a <n 且 0 <=b < m 且a的7进制和n-1的7进制位数相同 且b的7进制和m-1的7进制位数相同，还有a和b的7进制上的每位上的数各不相同。

看懂题目，就很简单了，先判断a和b的7进制位数是否超过7，不超过的话就dfs暴力枚举计算就可以了。

//#pragma comment(linker, "/STACK:102400000, 102400000")
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
typedef pair <int, int> P;
const int N = 1e5 + 5;
int num[10], n, m, ans;
bool vis[10];

int get(int num[], int len) {
    int res = 0, temp = 1;
    for(int i = len; i >= 1; --i) {
        res += num[len] * temp;
        temp *= 7;
    }
    return res;
}

void dfs(int len1, int len2, int dep) {
    if(dep == len1 + len2) {
        int sum = 0, temp = 1;
        for(int i = len1+len2; i >= len1+1; --i) {
            sum += temp*num[i];
            temp *= 7;
        }
        if(sum < m)
            ans++;
        return ;
    }
    else if(dep == len1) {
        int sum = 0, temp = 1;
        for(int i = len1; i >= 1; --i) {
            sum += temp*num[i];
            temp *= 7;
        }
        if(sum >= n)
            return ;
    }  
    dep++;
    for(int i = 0 ; i < 7 ; ++i) {
        if(vis[i])
            continue;
        vis[i] = true;
        num[dep] = i;
        dfs(len1, len2, dep);
        vis[i] = false;
    }

}

int main()
{
    cin >> n >> m;
    int cnt1 = 0, cnt2 = 0, temp = n - 1;
    do {
        temp /= 7;
        ++cnt1;
    } while(temp);
    temp = m - 1;
    do {
        temp /= 7;
        ++cnt2;
    } while(temp);
    if (cnt1 + cnt2 > 7) {
        cout << 0 << endl;
    } else {
        dfs(cnt1, cnt2, 0);
        cout << ans << endl;
    }
    return 0;
}