PAT 甲级 1011. World Cup Betting

原题传送门

• 就是找三组三个数中的最大值，按公式计算结果
• 注意：原题中的示例输出有误，实际AC的代码不需要将结果四舍五入的处理

代码一

#include <iostream>#include <algorithm>#include <iomanip>using namespace std;int main(int argc, const char * argv[]) {    double game1[3];    cin >> game1[0] >> game1[1] >> game1[2];    double max1 = *max_element(game1, game1 + 3);    double game2[3];    cin >> game2[0] >> game2[1] >> game2[2];    double max2 = *max_element(game2, game2 + 3);    double game3[3];    cin >> game3[0] >> game3[1] >> game3[2];    double max3 = *max_element(game3, game3 + 3);    if(max1 == game1[0])        cout << 'W' << ' ';    else if(max1 == game1[1])        cout << 'T' << ' ';    else        cout << 'L' << ' ';    if(max2 == game2[0])        cout << 'W' << ' ';    else if(max2 == game2[1])        cout << 'T' << ' ';    else        cout << 'L' << ' ';    if(max3 == game3[0])        cout << 'W' << ' ';    else if(max3 == game3[1])        cout << 'T' << ' ';    else        cout << 'L' << ' ';    cout << fixed << setprecision(2)         << (max1 * max2 * max3 * 0.65 - 1) * 2 << endl;    // 四舍五入保留两位小数的话，就把结果+0.005    return 0;}

代码二

#include <iostream>using namespace std;int main() {    int a[3];    char b[3]={'W','T','L'};    double sum=1,k;    int i,j;    for(i=0;i<3;i++) {        double max=0;        for(j=0;j<3;j++) {            scanf("%lf",&k);            if(k>max) {                max=k;                a[i]=j;            }        }        sum*=max;    }    printf("%c %c %c ",b[a[0]],b[a[1]],b[a[2]]);    printf("%.2f",(sum*0.65-1)*2);    return 0;  }

附原题

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 （应该是37.97）