Saturday, October 30, 2010


Problem Links:



World Cup

Time Limit: 1000MSMemory Limit: 65536K
Total Submissions: 5806Accepted: 3002


A World Cup of association football is being held with teams from around the world. The standing is based on the number of points won by the teams, and the distribution of points is done the usual way. That is, when a teams wins a match, it receives 3 points; if the match ends in a draw, both teams receive 1 point; and the loser doesn’t receive any points.

Given the current standing of the teams and the number of teams participating in the World Cup, your task is to determine how many matches ended in a draw till the moment.


The input contains several test cases. The first line of a test case contains two integers T and N, indicating respectively the number of participating teams (0 ≤ T ≤ 200) and the number of played matches (0 ≤ N ≤ 10000). Each one of the T lines below contains the name of the team (a string of at most 10 letter and digits), followed by a whitespace, then the number of points that the team obtained till the moment. The end of input is indicated by T = 0.


For each one of the test cases, your program should print a single line containing an integer, representing the quantity of matches that ended in a draw till the moment.

Sample Input
3 3
Brasil 3
Australia 3
Croacia 3
3 3
Brasil 5
Japao 1
Australia 1
0 0

Sample Output

South America 2006, Brazil Subregion


Easy straight forward.

Source Code:

[sourcecode language="cpp" collapse="true" padlinenumbers="true"]
//Tue May 25 00:56:40 CDT 2010
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

int main(int argc, const char* argv[])
// freopen("", "r", stdin);
// freopen("output.out", "w", stdout);
int T, N;
while (cin >> T >> N && T)
long long sum = 0;
string s;
int score;
for (int i = 0; i < T; i++)
cin >> s >> score;
sum += score;
cout << 3 * N - sum << endl;
// fclose(stdin);
// fclose(stdout);
return 0;

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