zoukankan      html  css  js  c++  java
  • 1011. World Cup Betting (20)

    With the 2010 FIFA World Cup running, football fans the world over were becoming increasingly excited as the best players from the best teams doing battles for the World Cup trophy in South Africa. Similarly, football betting fans were putting their money where their mouths were, by laying all manner of World Cup bets.

    Chinese Football Lottery provided a "Triple Winning" game. The rule of winning was simple: first select any three of the games. Then for each selected game, bet on one of the three possible results -- namely W for win, T for tie, and L for lose. There was an odd assigned to each result. The winner's odd would be the product of the three odds times 65%.

    For example, 3 games' odds are given as the following:

     W    T    L
    1.1  2.5  1.7
    1.2  3.0  1.6
    4.1  1.2  1.1
    

    To obtain the maximum profit, one must buy W for the 3rd game, T for the 2nd game, and T for the 1st game. If each bet takes 2 yuans, then the maximum profit would be (4.1*3.0*2.5*65%-1)*2 = 37.98 yuans (accurate up to 2 decimal places).

    Input

    Each input file contains one test case. Each case contains the betting information of 3 games. Each game occupies a line with three distinct odds corresponding to W, T and L.

    Output

    For each test case, print in one line the best bet of each game, and the maximum profit accurate up to 2 decimal places. The characters and the number must be separated by one space.

    Sample Input

    1.1 2.5 1.7
    1.2 3.0 1.6
    4.1 1.2 1.1
    

    Sample Output

    T T W 37.98
    #include "iostream"
    using namespace std;
    int main() {
        float a[3][3];
        int t[3];
        char p[3] = { 'W','T','L' };
        float sum = 1;
        for (int i = 0; i < 3; i++) {
            float MAX = 0;
            for (int j = 0; j < 3; j++) {
                cin >> a[i][j];
                if (a[i][j] > MAX) {
                    MAX = a[i][j];
                    t[i] = j;
                }
            }
            sum *= MAX;
        }
        for (int i = 0; i < 3; i++) {
            cout << p[t[i]] << " ";
        }
        printf("%.2f
    ", (sum * 0.65 - 1) * 2);
        return 0;
    }
  • 相关阅读:
    cri-o 与 cni的集成分析
    ocacle sql: 两张表左连接 ,1对多,取一条数据,取按时间最新的
    redis jedis pool 高并发的问题
    Redis事件订阅和持久化存储
    Spring缓存注解@Cacheable、@CacheEvict、@CachePut使用
    springboot配置redis
    Spring Boot使用Redis进行消息的发布订阅
    Spring boot中使用aop详解
    springBoot 全局异常捕捉
    pring boot中使用aop详解
  • 原文地址:https://www.cnblogs.com/minesweeper/p/6285656.html
Copyright © 2011-2022 走看看