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梦羽苍穹 — Mon, 10 May 2010 08:28:00 GMT

TimeLimit: 1 Second MemoryLimit: 32 Megabyte

Totalsubmit: 676 Accepted: 426

Description

Each employee of a bureaucracy has a job description - a few paragraphs that describe the responsibilities of the job. The employee's job description, combined with other factors, such as seniority, is used to determine his or her salary.

The Hay Point system frees the Human Resources department from having to make an intelligent judgement as to the value of the employee; the job description is merely scanned for words and phrases that indicate responsibility. In particular, job descriptions that indicate control over a large budget or management over a large number of people yield high Hay Point scores.

You are to implement a simplified Hay Point system. You will be given a Hay Point dictionary and a number of job descriptions. For each job description you are to compute the salary associated with the job, according to the system.

Input

The first line of input contains 2 positive integers: m <= 1000, the number of words in the Hay Point dictionary, and n <= 100, the number of job descriptions. m lines follow; each contains a word (a string of up to 16 lower-case letters) and a dollar value (a real number between 0 and 1,000,000). Following the dictionary are the n job descriptions. Each job description consists of one or more lines of text; for your convenience the text has been converted to lower case and has no characters other than letters, numbers, and spaces. Each job description is terminated by a line containing a period.

Output

For each job description, output the corresponding salary computed as the sum of the Hay Point values for all words that appear in the description. Words that do not appear in the dictionary have a value of 0.

Sample Input

7 2
administer 100000
spending 200000
manage 50000
responsibility 25000
expertise 100
skill 50
money 75000
the incumbent will administer the spending of kindergarden milk money
and exercise responsibility for making change he or she will share
responsibility for the task of managing the money with the assistant
whose skill and expertise shall ensure the successful spending exercise
.
this individual must have the skill to perform a heart transplant and
expertise in rocket science
.

Sample Output

700150
150
�q�题很水的，直接定义一string�c�L��l�，然后再从文段查找记录��p��Q�直接看代码

1#include<iostream>
2#include<string>
3using namespace std;
4struct{
5string s;
6int money;
7}dictionary[1000];
8string description[10000];
9int main()
10{
11    //freopen("in.txt","r",stdin);
12    //freopen("out.txt","w",stdout);
13    int m,n,temp,sum;
14    while(cin>>m>>n)
15    {
16                    for(int i=0;i<m;i++)
17                    cin>>dictionary[i].s>>dictionary[i].money;
18                    for(int i=0;i<n;i++)
19                    {
20                            sum=0;
21                            int j;
22                            for(j=0;;j++)
23                            {
24                                         cin>>description[j];
25                                         if(description[j]==".")
26                                         break;
27                            }
28                            temp=j;
29                            for(int i=0;i<m;i++)
30                            {
31                                    for(int j=0;j<temp;j++)
32                                    if(dictionary[i].s==description[j])
33                                    sum+=dictionary[i].money;
34                            }
35                            cout<<sum<<endl;
36                    }
37    }
38    //system("PAUSE");
39    return 0;
40}
41

梦羽苍穹 2010-05-10 16:28 发表评论

梦羽苍穹 — Mon, 10 May 2010 07:59:00 GMT

TimeLimit: 1 Second MemoryLimit: 32 Megabyte

Totalsubmit: 1413 Accepted: 326

Description

If a tree falls in the forest, and there's nobody there to hear, does it make a sound? This classic conundrum was coined by George Berkeley (1685-1753), the Bishop and influential Irish philosopher whose primary philosophical achievement is the advancement of what has come to be called subjective idealism. He wrote a number of works, of which the most widely-read are Treatise Concerning the Principles of Human Knowledge (1710) and Three Dialogues between Hylas and Philonous (1713) (Philonous, the "lover of the mind," representing Berkeley himself).

Input

A forest contains T trees numbered from 1 to T and P people numbered from 1 to P. Standard input consists of a line containing P and T followed by several lines, containing a pair of integers i and j, indicating that person i has heard tree j fall. People may have different opinions as to which trees, according to Berkeley, have made a sound.

Input contains multiple test cases. Subsequent test cases are separated with a single blank line.

Output

How many different opinions are represented in the input? Two people hold the same opinion only if they hear exactly the same set of trees. You may assume that P < 100 and T < 100.

Sample Input

3 4
1 2
3 3
1 3
2 2
3 2
2 4

Sample Output

1#include<stdio.h>
2#include<stdlib.h>
3int main()
4{
5int n,m,a,b,i,j,k;
6int s[100][100],arr[100];
7int flag,sum;
8char str[100];
9while(scanf("%d%d",&n,&m)!=EOF)
10{
11sum=0;
12for(i=0;i<n;i++)
13{
14arr[i]=0;
15for(j=0;j<m;j++)
16s[i][j]=0;
17}
18getchar();
19while (gets(str)!=NULL)
20{
21    if (str[0]=='\0')
22           break;
23      sscanf(str,"%d %d",&a,&b);
24    s[a-1][b-1]=1;
25}
26for(i=0;i<n;i++)
27{
28if(arr[i]==0)
29{
30sum++;
31for(j=i+1;j<n;j++)
32{
33flag=0;
34if(arr[j]==0)
35{
36for(k=0;k<m;k++)
37if(s[i][k]!=s[j][k])
38{
39flag=1;
40break;
41}
42if(flag==0)
43arr[j]=1;
44}
45}
46}
47}
48printf("%d\n",sum);
49}
50}

梦羽苍穹 2010-05-10 15:59 发表评论

梦羽苍穹 — Mon, 10 May 2010 07:18:00 GMT

TimeLimit : 1 Second Memorylimit : 32 Megabyte

Totalsubmit : 165 Accepted : 38

There is a football game between teamX and teamY, we assume the position of the center of football ground is (0, 0),the position of the teamY's goal is(100,0),the position of the teamX's goal is(-100,0).

Now xiaoA in teamX wants to pass the football to his teammate xiaoB, was xiaoB offside(��位)�Q?/span>

Input

The first line of input contains a positive integer N <= 10, the number of test cases. For each test case there is one line contains 24 number, the first and second number indicate the position of xiaoA, the third and forth number indicate the position of xiaoB, and the remaining numbers indicate the positions of the team members in teamY.

Output

If xiaoB is offside output "Oh, My god!", else output "Come on!"

Sample Input

-99 0 99 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0 10 0

Sample Output

Oh, My god!

Hint

If xiaoA passes the football to xiaoB�Q?/span> and there is no members of teamY in front of xiaoB�Q?/span>then xiaoB is offside.

�q�题很水�Q�但要注意越位规则，在彼方半场�ƈ且xiaob的x坐标大于彼方所有球员的x坐标��q��位。只需要判断teamY的所有队员的最大x坐标与teamX的xiaoB的坐标的关系卛_��?/p>

#include<iostream>
#include<cstdlib>
using namespace std;
int main()
{
    int ax,ay,bx,by,otherx,othery,max;//变量可重复��用，不过我觉得没必要
    int n;
    cin>>n;
    while(n--)
    {
              max=0;
              cin>>ax>>ay>>bx>>by;
              for(int i=0;i<10;i++)//求取teamy最大的X坐标
              {
                      cin>>otherx>>othery;
                      if(max<otherx)max=otherx;
              }
              if(bx>max)cout<<"Oh, My god!"<<endl;
              else
              cout<<"Come on!"<<endl;
    }
    //system("PAUSE");
    return 0;
}

梦羽苍穹 2010-05-10 15:18 发表评论

梦羽苍穹 — Mon, 10 May 2010 07:16:00 GMT

TimeLimit : 1 Second Memorylimit : 32 Megabyte

Totalsubmit : 954 Accepted : 214

xiaoA is a beautiful girl, and she has so many skirts that she won't wear a skirt a second time before the skirt is washed.Now xiaoA wants to go shopping,
how many skirts she can choose to wear.
In order to distinguish these skirts,she gives out every skirt with a unique name.

Input

The first line of input contains a positive integer T (T <= 100), the number of test cases.
Each test case contains two integers N and M (1<=N,M<=100) in the first line.
N is the number of skirts.M is the number of operations.
Then N lines follow.Every line contains a word and the word is not longer than 10.
Then M lines follow.Each line contains a character 'D' or 'W' and the name of a skirt.
'D' indicates that xiaoA has wore the skirt. 'W' indicates that xiaoA has washed the skirt.

Output

After the M operations, you need to output that how many skirts can be chosen by xiaoA to wear.

Sample Input

1
5 5
SkirtA
SkirtB
SkirtC
SkirtD
SkirtE
D SkirtA
D SkirtB
D SkirtC
D SkirtD
W SkirtA

Sample Output

水题一�?#8230;…�Q�把所有skirt name��d��C��个string数组中，然后��d��状态量�Q�同时与skirt name匚w��Q�再用一个bool行数�l�state 记录skirt的及(qi��ng)时状态�?/p>

1#include<iostream>
2#include<cstdlib>
3#include<string>
4using namespace std;
5string name[101];
6string  name_s;
7bool state[101];
8int main()
9{
10    int T,n,m,count;
11    char CH;
12    cin>>T;
13    while(T--)
14    {
15              count=0;
16              cin>>n>>m;
17              for(int i=0;i<n;i++)
18              state[i]=true;
19              for(int i=0;i<n;i++)
20              cin>>name[i];
21              for(int i=0;i<m;i++)
22              {
23                      cin>>CH;
24                      cin>>name_s;
25                      for(int j=0;j<n;j++)
26                      {
27                              if(name_s==name[j])
28                              {
29                                                 if(CH=='D')
30                                                 state[j]=false;
31                                                 else
32                                                 state[j]=true;
33                              }
34                      }
35              }
36              for(int i=0;i<n;i++)
37                      if(state[i])count++;
38              cout<<count<<endl;
39    }
40    //system("PAUSE");
41    return 0;
42}
43

梦羽苍穹 2010-05-10 15:16 发表评论

黑龙�?010�|�络赛解题报�?哈工�E?336 Losing Mobile Phone)

梦羽苍穹 — Mon, 10 May 2010 07:13:00 GMT

Losing Mobile Phone
TimeLimit : 1 Second Memorylimit : 32 Megabyte

Totalsubmit : 1 Accepted : 1

Description

XiaoA has N friends , now he would like to contact his friend N , but unfortunately , he lost his mobile phone, and he can just come up with friend 1 's phone number. Can you tell us whether he can build contact with his friend N.

Input

The first line of input contains a positive integer T <= 100, the number of test cases. For each case , there is an integer N (1<=N<=100) in the first line , means xiaoA has N friends , then follows N lines , in the ith line , there is an integer M first, means xiaoA’s friend i can contact with M xiaoA’s friends , then follows M integers kij , means friend i of xiaoA can contact with xiaoA’s friend kij (1<=i<=N, 1<=j<=M,1<=kij<=N) .

Output

For each test case , print one line "yes" if xiaoA can contact with friend N , or "no" instead.

Sample Input

1 2

1 3

1 4

1 5

4 1 2 3 4

3 2 3 4

3 1 3 4

3 1 2 4

3 1 2 3

4 1 2 3 4

Sample Output

yes

用递归的方法遍历所有friends�Q�查找是否会(x��)到达friends N如果到达则标记TRUE�?/p>

1#include<iostream>
2#include<cstdlib>
3using namespace std;
4void search(int n);
5int T,N;
6struct{
7       int relation[105];
8       bool sign;
9       int x;
10       }frien_d[105];
11int main()
12{
13    //freopen("in.dat","r",stdin);
14    //freopen("out.dat","w",stdout);
15    cin>>T;
16    while(T--)
17    {
18              cin>>N;
19              for(int i=1;i<=N;i++)
20              {
21                      cin>>frien_d[i].x;
22                      for(int j=1;j<=frien_d[i].x;j++)
23                      cin>>frien_d[i].relation[j];
24              }
25              for(int i=1;i<N;i++)
26              frien_d[i].sign=true;
27              frien_d[N].sign=false;
28              search(1);
29              if(frien_d[N].sign)
30              cout<<"yes"<<endl;
31              else
32              cout<<"no"<<endl;
33    }
34    //system("PAUSE");
35    return 0;
36}
37void search(int n)
38{
39      if(frien_d[n].sign==false)return;
40      for(int i=1;i<=frien_d[n].x;i++)
41      {
42              if(frien_d[n].relation[i]==N)
43              {
44                                           frien_d[N].sign=true;
45                                           return;
46                                           break;
47              }
48              else
49              {
50                   frien_d[n].sign=false;
51                   search(frien_d[n].relation[i]);
52              }
53      }
54}
55

梦羽苍穹 2010-05-10 15:13 发表评论