# 2016年微软笔试题目

Miss慧.java

### 描述

Steven loves reading book on his phone. The book he reads now consists of N paragraphs and the i-th paragraph contains ai characters.

Steven wants to make the characters easier to read, so he decides to increase the font size of characters. But the size of Steven's phone screen is limited. Its width is W and height is H. As a result, if the font size of characters is S then it can only show ⌊W / S⌋ characters in a line and ⌊H / S⌋ lines in a page. (⌊x⌋ is the largest integer no more than x)

So here's the question, if Steven wants to control the number of pages no more than P, what's the maximum font size he can set? Note that paragraphs must start in a new line and there is no empty line between paragraphs.

### 输入

Input may contain multiple test cases.

The first line is an integer TASKS, representing the number of test cases.

For each test case, the first line contains four integers N, P, W and H, as described above.

The second line contains N integers a1, a2, ... aN, indicating the number of characters in each paragraph.

For all test cases,

1 <= N <= 103,

1 <= W, H, ai <= 103,

1 <= P <= 106,

There is always a way to control the number of pages no more than P.

### 输出

For each testcase, output a line with an integer Ans, indicating the maximum font size Steven can set.

```2
1 10 4 3
10
2 10 4 3
10 10```

```3
2
```

### 描述

Little Hi runs a web server. Sometimes he has to deny access from a certain set of malicious IP addresses while his friends are still allow to access his server. To do this he writes N rules in the configuration file which look like:

```allow 1.2.3.4/30
deny 1.1.1.1
allow 127.0.0.1
allow 123.234.12.23/3
deny 0.0.0.0/0
```

When there comes a request, the rules are checked in sequence until the first match is found. If no rule is matched the request will be allowed. Rule and request are matched if the request address is the same as the rule address or they share the same first mask digits when both written as 32bit binary number.

For example IP "1.2.3.4" matches rule "allow 1.2.3.4" because the addresses are the same. And IP "128.127.8.125" matches rule "deny 128.127.4.100/20" because 10000000011111110000010001100100 (128.127.4.100 as binary number) shares the first 20 (mask) digits with10000000011111110000100001111101 (128.127.8.125 as binary number).

Now comes M access requests. Given their IP addresses, your task is to find out which ones are allowed and which ones are denied.

### 输入

Line 1: two integers N and M.

Line 2-N+1: one rule on each line.

Line N+2-N+M+1: one IP address on each line.

For 40% of the data: 1 <= N, M <= 1000.

For 100% of the data: 1 <= N, M <= 100000.

### 输出

For each request output "YES" or "NO" according to whether it is allowed.

```5 5
allow 1.2.3.4/30
deny 1.1.1.1
allow 127.0.0.1
allow 123.234.12.23/3
deny 0.0.0.0/0
1.2.3.4
1.2.3.5
1.1.1.1
100.100.100.100
219.142.53.100```

```YES
YES
NO
YES
NO
```

### 描述

You work as an intern at a robotics startup. Today is your company's demo day. During the demo your company's robot will be put in a maze and without any information about the maze, it should be able to find a way out.

The maze consists of N * M grids. Each grid is either empty(represented by '.') or blocked by an obstacle(represented by 'b'). The robot will be release at the top left corner and the exit is at the bottom right corner.

Unfortunately some sensors on the robot go crazy just before the demo starts. As a result, the robot can only repeats two operations alternatively: keep moving to the right until it can't and keep moving to the bottom until it can't. At the beginning, the robot keeps moving to the right.

```rrrrbb..
...r....     ====> The robot route with broken sensors is marked by 'r'.
...rrb..
...bb...
```

While the FTEs(full-time employees) are busy working on the sensors, you try to save the demo day by rearranging the maze in such a way that even with the broken sensors the robot can reach the exit successfully. You can change a grid from empty to blocked and vice versa. So as not to arouse suspision, you want to change as few grids as possible. What is the mininum number?

### 输入

Line 1: N, M.

Line 2-N+1: the N * M maze.

For 20% of the data, N * M <= 16.

For 50% of the data, 1 <= N, M <= 8.

For 100% of the data, 1<= N, M <= 100.

### 输出

The minimum number of grids to be changed.

```4 8
....bb..
........
.....b..
...bb...```

`1`

### 描述

Little Hi is playing a sandbox voxel game. In the game the whole world is constructed by massive 1x1x1 cubes. The edges of cubes are parallel to the coordinate axes and the coordinates (x, y, z) of the center of each cube are integers.

At the beginning there is nothing but plane ground in the world. The ground consists of all the cubes of z=0. Little Hi needs to build everything by placing cubes one by one following the rules:

1. The newly placed cube must be adjacent to the ground or a previously placed cube. Two cubes are adjacent if and only if they share a same face.

2. The newly placed cube must be accessible from outside which means by moving in 6 directions(up, down, left, right, forward, backward) there is a path from a very far place - say (1000, 1000, 1000) in this problem - to this cube without passing through ground or other cubes.

Given a sequence of cubes Little Hi wants to know if he can build the world by placing the cubes in such order.

### 输入

The first line contains the number of test cases T(1 <= T <= 10).

For each test case the first line is N the number of cubes in the sequence.

The following N lines each contain three integers x, y and z indicating the coordinates of a cube.

For 20% of the data, 1 <= N <= 1000, 1 <= x, y, z <= 10.

For 100% of the data, 1 <= N <= 100000, 1 <= x, y, z <= 100.

### 输出

For each testcase output "Yes" or "No" indicating if Little Hi can place the cubes in such order.

### 样例提示

In the first test case three cubes are placed on the ground. It's OK.

In the second test case (1, 3, 2) is neither on the ground nor adjacent to previous cubes. So it can't be placed.

In the last test case (2, 2, 1) can not be reached from outside. So it can't be placed.

```3
3
1 1 1
1 2 1
1 3 1
3
1 1 1
1 2 1
1 3 2
17
1 1 1
1 2 1
1 3 1
2 3 1
3 3 1
3 2 1
3 1 1
2 1 1
2 1 2
1 1 2
1 2 2
1 3 2
2 3 2
3 3 2
3 2 2
2 2 2
2 2 1```

```Yes
No
No```

# 11条回帖

• ```
第一题完成 是比较简单的！代码如下：

#include<iostream>
#include<stdio.h>
using namespace std;

int n,w,h,p;
int a[1010];

bool check (int s)
{
if (w<s || h<s) return 0;
int nowx,nowy;
int x=w/s;
int y=h/s;
int tot=0;
int pages=0;
for (int i=1;i<=n;i++)
{
tot+=a[i]/x;
if (a[i]%x) tot++;
while (tot>y)
{
tot-=y;
pages++;
}
if (pages>p) return 0;
}
if (tot) pages++;
if (pages<=p) return 1;
return 0;
}

void doing ()
{
cin>>n>>p>>w>>h;
for (int i=1;i<=n;i++)
cin>>a[i];
int left,right,ans;
left=0;right=200000000;
while (left<=right)
{
int mid=(left+right)>>1;
if (check(mid)){
ans=mid;
left=mid+1;
}
else right=mid-1;
}
cout<<ans<<endl;
}

int main ()
{
int T;
cin>>T;
while (T--)
doing ();
return 0;
}

```

发表于 2016-04-06 21:03:45 回复(0) 赞(0)
• outshine 2#

话说我也就完成了第一题，第二题本来想用个ip转换的库的，结果发现这是在线编程，并没有库给我用

发表于 2016-04-06 21:33:23 回复(2) 赞(0)
• jungieve 3#

总分加起来三位数多一点，sigh...还是刷题太少了
发表于 2016-04-06 21:37:25 回复(1) 赞(0)
• WAWAWA 4#

求最后一题思路。。。sigh 暴力骗了20分
发表于 2016-04-06 21:41:52 回复(9) 赞(0)
• 盛夏de午夜 5#

第一题20分钟过，第二题40，第三题没调处来，输出1骗了20。。。输出0也有10分，其实要是早点去骗，就去输出1或者0，至少能骗30分的
发表于 2016-04-06 21:44:22 回复(4) 赞(0)
• ght1993 6#

第三题动态规划吧
发表于 2016-04-06 22:39:40 回复(4) 赞(0)
• 第二题这样的，刚开始看错题目了耽误了好多时间！/(ㄒoㄒ)/~~
```#include<iostream>
#include<stdio.h>
#include<string>
#include<map>
using namespace std;

map<string, int>pp;
int n,m;

void work (string s)
{
bool p=0;
int now;
if (s[0]=='a') return ;
for (int i=5;i<s.size();i++)
if (s[i]=='/')
{
p=1;
now=i;
break;
}
string ss[5];
string temp="";
int count=1;
int last;
if (!p)
{
for (int i=5;i<s.size();i++)
{
if (s[i]!='.')
temp+=s[i];
if (s[i]=='.')
{
ss[count]=temp;
temp="";
count++;
last=i;
}
}
ss[4]="";
for (int i=last+1;i<s.size();i++)
ss[4]+=s[i];
}
else
{
for (int i=5;i<s.size();i++)
{
if (s[i]!='.')
temp+=s[i];
if (s[i]=='.')
{
ss[count]=temp;
temp="";
count++;
last=i;
}
}
ss[4]="";
for (int i=last+1;i<now;i++)
ss[4]+=s[i];
}
/*cout<<s<<endl;
cout<<ss[1]<<" ";
cout<<ss[2]<<" ";
cout<<ss[3]<<" ";
cout<<ss[4]<<" "<<endl;*/
int a[9],sz,x;
sz=0;
x=0;
for (int i=0;i<ss[1].size();i++)
x=x*10+(ss[1][i]-'0');
while (x)
{
a[++sz]=x%2;
x=x/2;
}
if (sz<8)
{
for (int i=1;i<=8-sz;i++)
temp+="0";
}
for (int i=sz;i>=1;i--)
temp+=char(a[i]+'0');
sz=0;
x=0;
for (int i=0;i<ss[2].size();i++)
x=x*10+(ss[2][i]-'0');
while (x)
{
a[++sz]=x%2;
x=x/2;
}
if (sz<8)
{
for (int i=1;i<=8-sz;i++)
temp+="0";
}
for (int i=sz;i>=1;i--)
temp+=char(a[i]+'0');
sz=0;
x=0;
for (int i=0;i<ss[3].size();i++)
x=x*10+(ss[3][i]-'0');
while (x)
{
a[++sz]=x%2;
x=x/2;
}
if (sz<8)
{
for (int i=1;i<=8-sz;i++)
temp+="0";
}
for (int i=sz;i>=1;i--)
temp+=char(a[i]+'0');
sz=0;
x=0;
for (int i=0;i<ss[4].size();i++)
x=x*10+(ss[4][i]-'0');
while (x)
{
a[++sz]=x%2;
x=x/2;
}
if (sz<8)
{
for (int i=1;i<=8-sz;i++)
temp+="0";
}
for (int i=sz;i>=1;i--)
temp+=char(a[i]+'0');
//cout<<temp<<endl;
//cout<<temp.size()<<endl;
string ans="";
for (int i=ss[4].size();i<temp.size();i++)
ans+=temp[i];
//cout<<ans<<endl;
//a[ans]=0;
string t="";
for (int i=0;i<=19;i++)
t+=ans[i];
pp[t]=1;
//cout<<ans<<endl;
//cout<<t<<endl;
}

void calc (string s)
{
bool p=0;
int now;
if (s[0]=='a') return ;
for (int i=0;i<s.size();i++)
if (s[i]=='/')
{
p=1;
now=i;
break;
}
string ss[5];
string temp="";
int count=1;
int last;
if (!p)
{
for (int i=0;i<s.size();i++)
{
if (s[i]!='.')
temp+=s[i];
if (s[i]=='.')
{
ss[count]=temp;
temp="";
count++;
last=i;
}
}
ss[4]="";
for (int i=last+1;i<s.size();i++)
ss[4]+=s[i];
}
else
{
for (int i=0;i<s.size();i++)
{
if (s[i]!='.')
temp+=s[i];
if (s[i]=='.')
{
ss[count]=temp;
temp="";
count++;
last=i;
}
}
ss[4]="";
for (int i=last+1;i<now;i++)
ss[4]+=s[i];
}
/*cout<<s<<endl;
cout<<ss[1]<<" ";
cout<<ss[2]<<" ";
cout<<ss[3]<<" ";
cout<<ss[4]<<" "<<endl;*/
int a[9],sz,x;
sz=0;
x=0;
for (int i=0;i<ss[1].size();i++)
x=x*10+(ss[1][i]-'0');
while (x)
{
a[++sz]=x%2;
x=x/2;
}
if (sz<8)
{
for (int i=1;i<=8-sz;i++)
temp+="0";
}
for (int i=sz;i>=1;i--)
temp+=char(a[i]+'0');
sz=0;
x=0;
for (int i=0;i<ss[2].size();i++)
x=x*10+(ss[2][i]-'0');
while (x)
{
a[++sz]=x%2;
x=x/2;
}
if (sz<8)
{
for (int i=1;i<=8-sz;i++)
temp+="0";
}
for (int i=sz;i>=1;i--)
temp+=char(a[i]+'0');
sz=0;
x=0;
for (int i=0;i<ss[3].size();i++)
x=x*10+(ss[3][i]-'0');
while (x)
{
a[++sz]=x%2;
x=x/2;
}
if (sz<8)
{
for (int i=1;i<=8-sz;i++)
temp+="0";
}
for (int i=sz;i>=1;i--)
temp+=char(a[i]+'0');
sz=0;
x=0;
for (int i=0;i<ss[4].size();i++)
x=x*10+(ss[4][i]-'0');
while (x)
{
a[++sz]=x%2;
x=x/2;
}
if (sz<8)
{
for (int i=1;i<=8-sz;i++)
temp+="0";
}
for (int i=sz;i>=1;i--)
temp+=char(a[i]+'0');
//cout<<temp<<endl;
//cout<<temp.size()<<endl;
string ans="";
for (int i=ss[4].size();i<temp.size();i++)
ans+=temp[i];
//cout<<ans<<endl;
//a[ans]=0;
string t="";
for (int i=0;i<=19;i++)
t+=ans[i];
//cout<<t<<endl;
if (pp[t]) cout<<"NO"<<endl;
else cout<<"YES"<<endl;
}

void doing ()
{
cin>>n>>m;
getchar();
string s;
for (int i=1;i<=n;i++)
{
getline (cin,s);
work (s);
}
//cout<<"!!!!!"<<pp["00000001000000010000"];
for (int i=1;i<=m;i++)
{
cin>>s;
calc (s);
}
}

int main ()
{
doing ();
return 0;
}```

发表于 2016-04-07 08:57:41 回复(5) 赞(0)
• 一梦一程 8#

AC一道.....
发表于 2016-04-07 15:21:31 回复(1) 赞(0)
• ```#include<algorithm>
#include<utility>
#include<string>
#include<sstream>
#include<iostream>
#include<vector>
#include<map>
#include<bitset>
using namespace std;
replace(str.begin(), str.end(), '.', ' ');
replace(str.begin(), str.end(), '/', ' ');
istringstream istr(str);
pair<unsigned, unsigned> pa;
for (int i{},j;i < 4;++i) {
istr >> j;
pa.first += j*(1 << 8 * (3 - i));
}
!istr.eof() ? istr >> pa.second : (pa.second = 32, istr);
return pa;
}
void trim_front(std::string& str, char ch) {
size_t sz = str.find_first_not_of(ch, 0);
str.erase(0, sz);
}
class search_tree {
public:
~search_tree();
private:
enum class states
{
mid_state ,allow, deny
};
struct node
{
node *left, *right;
states acc;
int num;
node() :left{ nullptr }, right{ nullptr }, acc{ states::mid_state } {}
};
void del(node *p);
};
using namespace std;
for (int i{};i < mask;++i) {
if (pos->acc != states::mid_state)
return;
if (bit[31-i]) {
if (pos->right == nullptr)
pos->right = new node{};
pos = pos->right;
}
else
{
if (pos->left == nullptr)
pos->left = new node{};
pos = pos->left;
}
}
if (pos->acc != states::mid_state) return;
pos->num = num;
pos->acc = acc ? states::allow : states::deny;
}
using namespace std;
node * pos = head, *minnum{ nullptr };
int num=numeric_limits<int>::max();
for (size_t i = 0;i<32&&pos; i++)
{
if (pos&&pos->acc != states::mid_state) {
if (pos->num < num) {
num = pos->num;
minnum = pos;
}

}
pos = bit[31-i] ?  pos->right : pos->left;

}
if (minnum == nullptr) return true;
return minnum->acc == states::allow ? true : false;

}
search_tree::~search_tree() {
}
void search_tree::del(search_tree::node* p) {
if (p->left)
del(p->left);
if (p->right)
del(p->right);
delete p;
}
int main() {
using namespace std;
for (int i, j;cin >> i >> j;) {
string str;
search_tree treenet;
while (isspace(cin.peek()))
{
cin.get();
}
for (int k{};k < i;++k) {
getline(cin, str);
trim_front(str, ' ');
auto pos = str.find(' ');
auto ps = getad_ms(string(str, pos + 1));
bool acc = str[0] == 'a' ? true:false;
}
for (int k{};k < j;++k) {
cin >> str;
cout << (treenet.accept(pa.first)?"YES":"NO") << '\n';
}
}
}```

发表于 2016-04-09 00:18:29 回复(1) 赞(0)
• 三更雨123456 10#

各位大神们，可以用本地IDE吗？还是必须在线编？
还有面试需要准备英文自我介绍之类的吗
发表于 2017-03-30 13:40:50 回复(0) 赞(0)
• 苟有 11#

```//第二题详解
#include<iostream>
#include<string>
#include<vector>
using namespace std;

struct IPallow{
int flag;  //1 allow or 0 deny;
vector<int> ip;
};

vector<int> IpStrToIpInt(string ipstr)
{
vector<int> ipint;
int index=0;
while(index!=string::npos)
{
index=ipstr.find('.');
if(index==string::npos) {ipint.push_back(atoi(ipstr.c_str())); break;}
ipint.push_back(atoi((ipstr.substr(0,index)).c_str()));
ipstr=ipstr.substr(index+1,ipstr.length()-index-1);
}

return ipint;
}

vector<IPallow> StringtoIPallow(vector<string> list)
{
vector<IPallow> iplist;
IPallow ip;
int space,slash;
for(int i=0;i<list.size();i++)
{
temp=list[i];
space=temp.find(' ');
allow=temp.substr(0,space);
right=temp.substr(space+1,temp.length()-space);
slash=right.find('/');
if(slash!=string::npos)
{
ipstr=right.substr(0,slash);
}
else
{
ipstr=right;
}
if(allow=="allow") ip.flag=1;
else ip.flag=0;
ip.ip=IpStrToIpInt(ipstr);
iplist.push_back(ip);
}

return iplist;
}

void JudgeIP(vector<string> list,string q)
{
vector<int> qin=IpStrToIpInt(q);
vector<IPallow> iplist=StringtoIPallow(list);
vector<int> temp;
IPallow tempip;
for(int i=0;i<iplist.size();i++)
{
temp=qin;
tempip=iplist[i];
{
if(tempip.ip==temp) {
if(tempip.flag)
cout<<"Yes"<<endl;
else cout<<"No"<<endl;
return;
}
continue;
}
{
tempip.ip[3]=0;  temp[3]=0;
if(tempip.ip==temp) {
if(tempip.flag)
cout<<"Yes"<<endl;
else cout<<"No"<<endl;
return;
}
continue;
}
{
tempip.ip[3]=0;  temp[3]=0;
tempip.ip[2]=0;  temp[2]=0;
if(tempip.ip==temp) {
if(tempip.flag)
cout<<"Yes"<<endl;
else cout<<"No"<<endl;
return;
}
continue;
}
{
tempip.ip[3]=0; temp[3]=0;
tempip.ip[2]=0; temp[2]=0;
tempip.ip[1]=0; temp[1]=0;
if(tempip.ip==temp) {
if(tempip.flag)
cout<<"Yes"<<endl;
else cout<<"No"<<endl;
return;
}
continue;
}
}
cout<<"No"<<endl;
return;
}

int main(){
int n,m;
cin>>n>>m;
vector<string> list,qu;
string temp;
getchar();
for(int i=0;i<n;i++){ getline(cin,temp); list.push_back(temp);}
for(int i=0;i<m;i++){ getline(cin,temp); qu.push_back(temp);}

for(int i=0;i<m;i++){ JudgeIP(list,qu[i]);}
system("pause");
return 0;
}```

发表于 2017-03-31 13:49:40 回复(0) 赞(0)

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