codeforces 686C C. Robbers' watch(dfs)
题目链接:
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 题意: 每天n小时,每小时m分钟,现在给你一个7进制的表,问出现的所有的时间中数字全不相同的时间有多少个; 思路: 先找出表上有多少位数字,再按位dfs,看最后得到的数是否<n和<m,计数就好; AC代码:
//#include <bits/stdc++.h>
#include <vector>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <cstring>
#include <algorithm>
#include <cstdio> using namespace std;
#define Riep(n) for(int i=1;i<=n;i++)
#define Riop(n) for(int i=0;i<n;i++)
#define Rjep(n) for(int j=1;j<=n;j++)
#define Rjop(n) for(int j=0;j<n;j++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<''||CH>'';F= CH=='-',CH=getchar());
for(num=;CH>=''&&CH<='';num=num*+CH-'',CH=getchar());
F && (num=-num);
}
int stk[], tp;
template<class T> inline void print(T p) {
if(!p) { puts(""); return; }
while(p) stk[++ tp] = p%, p/=;
while(tp) putchar(stk[tp--] + '');
putchar('\n');
} const LL mod=1e9+;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=(<<)+;
const int maxn=; LL n,m,ans=;
int a[],b[],cnt1=,cnt2=,vis[];
LL fn,fm;
LL p[]; void DFS2(int po,LL num)
{
if(po>cnt2)
{
if(num<fm)ans++;
return ;
}
for(int i=;i<;i++)
{
if(!vis[i])
{
vis[i]=;
DFS2(po+,num+(LL)i*p[po]);
vis[i]=;
}
} } void DFS1(int po,LL num)
{
if(po>cnt1)
{
if(num<fn){DFS2(,);}
return ;
}
for(int i=;i<;i++)
{
if(!vis[i])
{
vis[i]=;
DFS1(po+,num+(LL)i*p[po]);
vis[i]=;
}
}
} int main()
{
read(n);read(m);
fn=n,fm=m;
if(n>)n--;
if(m>)m--;
LL w=;
for(int i=;i<;i++)
{
p[i]=w;
w=w*;
}
while(n)
{
a[++cnt1]=n%;
n/=;
}
while(m)
{
b[++cnt2]=m%;
m/=;
}
mst(vis,);
ans=;
DFS1(,);
cout<<ans<<"\n";
return ;
}
codeforces 686C C. Robbers' watch(dfs)的更多相关文章
- codeforces 615 B. Longtail Hedgehog (DFS + 剪枝)
题目链接: codeforces 615 B. Longtail Hedgehog (DFS + 剪枝) 题目描述: 给定n个点m条无向边的图,设一条节点递增的链末尾节点为u,链上点的个数为P,则该链 ...
- codeforces 711D Directed Roads(DFS)
题目链接:http://codeforces.com/problemset/problem/711/D 思路:由于每个点出度都为1,所以没有复杂的环中带环.DFS遍历,若为环则有2^k-2种,若为链则 ...
- Codeforces 600 E. Lomsat gelral (dfs启发式合并map)
题目链接:http://codeforces.com/contest/600/problem/E 给你一棵树,告诉你每个节点的颜色,问你以每个节点为根的子树中出现颜色次数最多的颜色编号和是多少. 最容 ...
- Codeforces 711 D. Directed Roads (DFS判环)
题目链接:http://codeforces.com/problemset/problem/711/D 给你一个n个节点n条边的有向图,可以把一条边反向,现在问有多少种方式可以使这个图没有环. 每个连 ...
- Vasya and a Tree CodeForces - 1076E(线段树+dfs)
I - Vasya and a Tree CodeForces - 1076E 其实参考完别人的思路,写完程序交上去,还是没理解啥意思..昨晚再仔细想了想.终于弄明白了(有可能不对 题意是有一棵树n个 ...
- Codeforces 931D Peculiar apple-tree(dfs+思维)
题目链接:http://codeforces.com/contest/931/problem/D 题目大意:给你一颗树,每个节点都会长苹果,然后每一秒钟,苹果往下滚一个.两个两个会抵消苹果.问最后在根 ...
- Codeforces 375D - Tree and Queries(dfs序+莫队)
题目链接:http://codeforces.com/contest/351/problem/D 题目大意:n个数,col[i]对应第i个数的颜色,并给你他们之间的树形关系(以1为根),有m次询问,每 ...
- Codeforces 667C Reberland Linguistics【DFS】
一道卡题意的题. 题目链接: http://codeforces.com/problemset/problem/667/C 题意: 一个串可以看成一个长度大于4的根,加上其后面的若干个相邻(in a ...
- Codeforces 659E New Reform【DFS】
题目链接: http://codeforces.com/problemset/problem/659/E 题意: 给定n个点和m条双向边,将双向边改为单向边,问无法到达的顶点最少有多少个? 分析: 无 ...
随机推荐
- CentOS7 设置代理
大多数公司的网络都使用局域网加代理上网,也就是说上外网必须使用公司指定的代理服务器,这有几个好处: 1. 首先代理可以一定程度提高浏览速度,因为可以将更多的网页缓存在代理服务器上,需要的时候直接拿就很 ...
- laravel 文件删除
删除文件 <?php class demo{ public function del(){ $disk = Storage::disk('public');//获取磁盘实例 $disk-> ...
- 关于srand()rand()的用法
转自:http://baike.baidu.com/link?url=bhos65ZKp8lEq_6chSsmQv29jHrqjN_IFGVMNod6BuicQ-3oCP5VsEn3RBjXBPvA7 ...
- 【2018 Multi-University Training Contest 2】
01: 02: 03: 04:https://www.cnblogs.com/myx12345/p/9394511.html 05: 06: 07:https://www.cnblogs.com/my ...
- 从零开始写STL—functional
function C++11 将任意类型的可调用(Callable)对象与函数调用的特征封装到一起. 这里的类是对函数策略的封装,将函数的性质抽象成组件,便于和algorithm库配合使用 基本运算符 ...
- T2627 村村通 codevs
http://codevs.cn/problem/2627/ 时间限制: 1 s 空间限制: 32000 KB 题目等级 : 黄金 Gold 题目描述 Description 农民约翰被选为他们 ...
- 洛谷——P2916 [USACO08NOV]为母牛欢呼Cheering up the Cows
https://www.luogu.org/problem/show?pid=2916 题目描述 Farmer John has grown so lazy that he no longer wan ...
- 【SDCC讲师专访】PingCAP联合创始人兼CEO刘奇:好的产品应开源,不闭门造车-CSDN.NET
[SDCC讲师专访]PingCAP联合创始人兼CEO刘奇:好的产品应开源,不闭门造车-CSDN.NET 小米的Themis
- 使用python分析解压zip、jar包等
python内置zipfile import zipfile, os zipFile = zipfile.ZipFile(os.path.join(os.getcwd(), 'txt.zip')) f ...
- CV_HAAR_FEATURE_DESC_MAX和CV_HAAR_FEATURE_MAX
#define CV_HAAR_FEATURE_MAX 3 //提前定义的一个宏,在程序中表示一个haar特征由至多三个矩形组成 #define CV_HAAR_FEATURE_DESC_MAX 20 ...