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http://codeforces.com/contest/493

A 第一次结构体开二维数组。。。

A. Vasya and Football

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya has started watching football games. He has learned that for some fouls
the players receive yellow cards, and for some fouls they receive red cards. A
player who receives the second yellow card automatically receives a red card.

Vasya is watching a recorded football match now and makes notes of all the
fouls that he would give a card for. Help Vasya determine all the moments in
time when players would be given red cards if Vasya were the judge. For each
player, Vasya wants to know only the first moment of time when he would
receive a red card from Vasya.

Input

The first line contains the name of the team playing at home. The second line
contains the name of the team playing away. Both lines are not empty. The
lengths of both lines do not exceed 20. Each line contains only of large
English letters. The names of the teams are distinct.

Next follows number _n_ ( 1 ≤ _n_ ≤ 90 ) — the number of fouls.

Each of the following _n_ lines contains information about a foul in the
following form:

  • first goes number _t_ ( 1 ≤ _t_ ≤ 90 ) — the minute when the foul occurs;
  • then goes letter “ h “ or letter “ a “ — if the letter is “ h “, then the card was given to a home team player, otherwise the card was given to an away team player;
  • then goes the player’s number _m_ ( 1 ≤ _m_ ≤ 99 );
  • then goes letter “ y “ or letter “ r “ — if the letter is “ y “, that means that the yellow card was given, otherwise the red card was given.

The players from different teams can have the same number. The players within
one team have distinct numbers. The fouls go chronologically, no two fouls
happened at the same minute.

Output

For each event when a player received his first red card in a chronological
order print a string containing the following information:

  • The name of the team to which the player belongs;
  • the player’s number in his team;
  • the minute when he received the card.

If no player received a card, then you do not need to print anything.

It is possible case that the program will not print anything to the output (if
there were no red cards).

Sample test(s)

input

MC
CSKA
9
28 a 3 y
62 h 25 y
66 h 42 y
70 h 25 y
77 a 4 y
79 a 25 y
82 h 42 r
89 h 16 y
90 a 13 r

output

MC 25 70
MC 42 82
CSKA 13 90


#include<stdio.h>  
#include<iostream>  
#include<math.h>  
#include<stdlib.h>  
#include<ctype.h>  
#include<algorithm>  
#include<vector>  
#include<string.h>  
#include<queue>  
#include<stack>  
#include<set>  
#include<map>  
#include<sstream>  
#include<time.h>  
#include<utility>  
#include<malloc.h>  
#include<stdexcept>  

using namespace std;  

char zhudui[25];
char kedui[25];

int n;

struct 
{
    char dui[25];
    int num;
    int vis;
    int p;
}p[2][100];

int T,NUM;
char DUI[2],P[2];

int main ()
{
    while (scanf("%s",zhudui)!=EOF)
    {
        scanf("%s",kedui);
        scanf("%d",&n);

        for(int i=0;i<=99;i++)
        {
            p[0][i].p = p[1][i].p = 0;
            p[1][i].vis = p[0][i].vis = 0;
        }

        for(int i=1;i<=n;i++)
        {
            cin>>T>>DUI>>NUM>>P;
            if (DUI[0] == 'h')
            {
                strcpy (p[0][NUM].dui , zhudui);

                if (P[0]=='y')
                    p[0][NUM].p+=1;
                else if (P[0] == 'r')
                    p[0][NUM].p+=2;

                if (p[0][NUM].p>=2 && p[0][NUM].vis == 0)
                {
                    cout<<p[0][NUM].dui<<" "<<NUM<<" "<<T<<endl;
                    p[0][NUM].vis = 1;
                }
            }
            else if (DUI[0] == 'a')
            {
                strcpy (p[1][NUM].dui , kedui);

                if (P[0]=='y')
                    p[1][NUM].p+=1;
                else if (P[0] == 'r')
                    p[1][NUM].p+=2;

                if (p[1][NUM].p>=2 && p[1][NUM].vis == 0)
                {
                    cout<<p[1][NUM].dui<<" "<<NUM<<" "<<T<<endl;
                    p[1][NUM].vis = 1;
                }
            }
        }

    }
    return 0;
}

B 简单题

B. Vasya and Wrestling

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya has become interested in wrestling. In wrestling wrestlers use
techniques for which they are awarded points by judges. The wrestler who gets
the most points wins.

When the numbers of points of both wrestlers are equal, the wrestler whose
sequence of points is lexicographically greater , wins.

If the sequences of the awarded points coincide, the wrestler who performed
the last technique wins. Your task is to determine which wrestler won.

Input

The first line contains number _n_ — the number of techniques that the
wrestlers have used ( 1 ≤ _n_ ≤ 2·10 5 ).

The following _n_ lines contain integer numbers _a_ _i_ ( | _a_ _i_ | ≤ 10
9 , _a_ _i_ ≠ 0 ). If _a_ _i_ is positive, that means that the first
wrestler performed the technique that was awarded with _a_ _i_ points. And if
_a_ _i_ is negative, that means that the second wrestler performed the
technique that was awarded with ( - _a_ _i_ ) points.

The techniques are given in chronological order.

Output

If the first wrestler wins, print string “ first “, otherwise print “
second “

Sample test(s)

input

5
1
2
-3
-4
3

output

second

input

3
-1
-2
3

output

first

input

2
4
-4

output

second

Note

Sequence _x_ = _x_ 1 _x_ 2 … _x_ | _x_ | is lexicographically larger
than sequence _y_ = _y_ 1 _y_ 2 … _y_ | _y_ | , if either | _x_ |   >
| _y_ | and _x_ 1 = _y_ 1 , _x_ 2 = _y_ 2 , … , _x_ | _y_ | = _y_ |
_y_ | , or there is such number _r_ ( _r_ <  | _x_ | , _r_ <  | _y_ | ),
that _x_ 1 = _y_ 1 , _x_ 2 = _y_ 2 ,  … , _x_ _r_ = _y_ _r_ and _x_
_r_ +  1 > _y_ _r_ +  1 .

We use notation | _a_ | to denote length of sequence _a_ .

#include<stdio.h>  
#include<iostream>  
#include<math.h>  
#include<stdlib.h>  
#include<ctype.h>  
#include<algorithm>  
#include<vector>  
#include<string.h>  
#include<queue>  
#include<stack>  
#include<set>  
#include<map>  
#include<sstream>
#include<time.h>  
#include<utility>  
#include<malloc.h>  
#include<stdexcept>  

using namespace std;  

long long  a[200010];

int main()
{
    int n;
    while (cin >> n)
    {
        long long  sum = 0;
        vector<int> c, d;

        for (int i = 0; i < n; i++) 
        {
            cin >> a[i];
            sum += a[i];
            if (a[i] > 0)
                c.push_back(a[i]);
            else
                d.push_back(-a[i]); 
        }

        if (sum != 0) 
        {
            if (sum > 0)
                cout << "first" << endl;
            else
                cout << "second" << endl;
            continue;       
        }

        int l = min(c.size(), d.size());

        bool flag = true;
        for (int i = 0; i < l; i++) 
        {
            if (c[i] == d[i])
                continue;
            if (c[i] < d[i]) {
                cout << "second" << endl;      
            }
            else {
                cout << "first" << endl;
            }
            flag = false;
            break;
        }
        if (!flag)
            continue;
        if (c.size() == d.size()) 
        {
            if (a[n-1] > 0)
                cout << "first" << endl;
            else
                cout << "second" << endl;             
        }
        else 
        {
            if (c.size() > d.size())
                cout << "first" << endl;
            else
                cout << "second" << endl;     
        }

    }
    return 0;
}