PAT甲级1011 World Cup Betting

最新推荐文章于 2025-08-10 13:27:23 发布

倾鸢染

最新推荐文章于 2025-08-10 13:27:23 发布

阅读量240

点赞数

CC 4.0 BY-SA版权

分类专栏：刷题题解文章标签： PAT 甲级 1011 World Cup Betting

本文链接：https://blog.youkuaiyun.com/qq_41895427/article/details/99693520

刷题题解专栏收录该内容

23 篇文章

订阅专栏

本文解析了PAT甲级1011题目的核心算法，通过遍历三维数组，寻找最佳投注组合以实现最大利润。示例展示了如何在三场比赛中选择投注类型(W、T、L)，并通过特定公式计算最大收益。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

PAT甲级1011

题目
1011 World Cup Betting (20 分)
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 3.1 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 (4.1×3.1×2.5×65%−1)×2=39.31 yuans (accurate up to 2 decimal places).

Input Specification:
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 Specification:
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 3.1 1.6
4.1 1.2 1.1
Sample Output:
T T W 39.31
题目意思
有三个比赛，每个比赛下注的赔率已经告诉你，问你最大的利润。

思路
用二维数组遍历一遍，选出每次比赛赔率最高的并记录状态(即W、T、L)最后根据公式算出最大利润。
输出要注意一下，先输出状态，然后输出最大利润，之间都有空格。

#include<cstdio>
#i
nclude<cstring>
#include<iostream>
#include<algorithm>

using namespace std;

int main()
{
    double hx[3][3],sum=1.0,mz=0.0;
    for(int i=0; i<3; i++)
    {
        for(int j=0; j<3; j++)
        {
            scanf("%lf",&hx[i][j]);
        }
    }
    char a[5];
    for(int i=0; i<3; i++)
    {
        for(int j=0; j<3; j++)
        {
            if(hx[i][j]>mz)
            {
                mz=hx[i][j];
                if(j==0)
                    a[i] = 'W';
                else if(j==1)
                    a[i] = 'T';
                else
                    a[i] = 'L';
            }

        }
        sum = sum*mz;
        mz = 0.0;
    }
    for(int i=0; i<3; i++)
        printf("%c ",a[i]);
    sum = (sum*0.65-1)*2;
    printf("%.2lf\n",sum);
    return 0;
}