2016-2017 ACM-ICPC Asia China-Final E-bet

2016 ACM-ICPC Asia China-Final E-bet

Description

The Codejamon game is on re! Fans across the world are predicting and betting on which team
will win the game.

A gambling company is providing betting odds for all teams; the odds for the ith team is Ai:Bi.
For each team, you can bet any positive amount of money, and you do not have to bet the same
amount on each team. If the ith team wins, you get your bet on that team back, plus Bi/Ai times
your bet on that team.

For example, suppose that there are two teams, with odds of 5:3 and 2:7 and you bet $20 on the
first team and $10 on the second team. If the fi rst team wins, you will lose your $10 bet on the
second team, but you will receive your $20 bet back, plus 3/5* 20 = 12, so you will have a total
of $32 at the end. If the second team wins, you will lose your $20 bet on the first team, but you
will receive your $10 bet back, plus 7/2 *10 = 35, so you will have a total of $45 at the end. Either
way, you will have more money than you bet ($20+$10=$30).

As a greedy fan, you want to bet on as many teams as possible to make sure that as long as one
of them wins, you will always end up with more money than you bet. Can you figure out how
many teams you can bet on?

Input

The input starts with one line containing exactly one integer T, which is the number of test cases.
Each test case starts with one line containing an integer N: the number of teams in the game.
Then, N more lines follow. Each line is a pair of numbers in the form Ai:Bi (that is, a number
Ai, followed by a colon, then a number Bi, with no spaces in between), indicating the odds for
the ith team.

Output

For each test case, output one line containing “Case #x: y”, where x is the test case number
(starting from 1) and y is the maximum number of teams that you can bet on, under the conditions
speci ed in the problem statement.

Examples

Input

1
3
1:1.1
1:0.2
1.5:1.7

Output

Case #1: 2

Problem description

有n个队伍参加比赛，给你赔率比，输出你能投入钱的最大队伍的个数。在所有你投入的队伍中，要保证只要任何一支队伍赢，所获得的金钱大于你所投入的总金钱。（各个队伍之间独立）

Solution

>设pi为投入第i个队伍的金钱占总金钱的比例> > 设 p i 为 投 入 第 i 个 队 伍 的 金 钱 占 总 金 钱 的 比 例 >

>则pi∗（1+Bi/Ai）>1> > 则 p i ∗ （ 1 + B i / A i ） > 1 >

>即pi>Ai/(Ai+Bi)> > 即 p i > A i / ( A i + B i ) >

则可按照

>Ai/(Ai+Bi)> > A i / ( A i + B i ) >

从小到大排序，只要总和小于1，则从小到大一直选择。

Code

/*
 * @Author: Simon 
 * @Date: 2018-08-13 21:49:19 
 * @Last Modified by: Simon
 * @Last Modified time: 2018-08-13 21:58:37
 */
#include 
using namespace std;
typedef int Int;
#define int long long
#define INF 0x3f3f3f3f
#define maxn 100005
long double a[maxn];
const int esp = 1000;
Int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    for (int ca = 1; ca <= t; ca++)
    {
        cout << "Case #" << ca << ": ";
        int n;
        cin >> n;
        for (int i = 1; i <= n; i++)
        {
            long double A, B;
            char c;
            cin >> A >> c >> B;
            int t1 = (int)(A * esp + 0.01);
            int t2 = (int)(B * esp + 0.01);
            a[i] = 1.0 * A / (long double)(A + B);
        }
        sort(a + 1, a + n + 1);
        long double sum = 0;
        int ans = 0;
        for (int i = 1; i <= n; i++)
        {
            if (sum + a[i] >= 1)
                break;
            ans++;
            sum += a[i];
        }
        cout << ans << endl;
    }
    cin.get(), cin.get();
    return 0;
}