1011. World Cup Betting (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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    L1.1  2.5  1.71.2  3.0  1.64.1  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 1.71.2 3.0 1.64.1 1.2 1.1

Sample Output

T T W 37.98

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这道题的理解：1.数据分为3组看（以行划分，不是以列划分），因为一行就是一次match

2.其实可以很简单的考虑char数组，然后只要记录下标来输出。。。但是我就是想要用map。。。然后改了很久才过测试点

3.sort的用法………

#include<cstdio>#include<map>#include<algorithm>using namespace std;map <float, int> m;  //map取名字不能取map!!!bool compare(float a, float b){return a>b;   //升序排列，如果改为return a>b，则为降序}int main() {float a[3];float b[3];for (int j = 0; j <3; j++) {for (int i = 0; i < 3; i++) {scanf("%f", &a[i]);//printf("66666 %f\n", a[i*3 + j]);m.insert(pair<float, int>(a[i], i));}sort(a, a +3, compare);  //sort参数是a和a+3,这样才是将三个数排序，以前错的就是写a+2，所以会有几个测试点过不去//printf("3333 %lf \n", a[0]);b[j] = a[0];if (m[a[0]] == 0) {           //一定要把判定放在每次的循环里，不然下一次可能会有相同的小数但是对应不同的整数位置覆盖这一次的map数据//ans[j] = { 'W' };printf("W ");}else if (m[a[0]] == 1) {//ans[j] = { 'T' };printf("T ");}else {//ans[j] = { 'L' };printf("L ");}}printf("%.2f\n", ((b[0] * b[1] * b[2])*0.65 - 1) * 2);return 0;}