Rock-Paper-Scissors Tournament
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2178 Accepted Submission(s): 693

Problem Description
Rock-Paper-Scissors is game for two players, A and B, who each choose, independently of the other, one of rock, paper, or scissors. A player chosing paper wins over a player chosing rock; a player chosing scissors wins over a player chosing paper; a player chosing rock wins over a player chosing scissors. A player chosing the same thing as the other player neither wins nor loses.
A tournament has been organized in which each of n players plays k rock-scissors-paper games with each of the other players - k games in total. Your job is to compute the win average for each player, defined as w / (w + l) where w is the number of games won, and l is the number of games lost, by the player.

Input
Input consists of several test cases. The first line of input for each case contains 1 ≤ n ≤ 100 1 ≤ k ≤ 100 as defined above. For each game, a line follows containing p1, m1, p2, m2. 1 ≤ p1 ≤ n and 1 ≤ p2 ≤ n are distinct integers identifying two players; m1 and m2 are their respective moves ("rock", "scissors", or "paper"). A line containing 0 follows the last test case.

Output
Output one line each for player 1, player 2, and so on, through player n, giving the player's win average rounded to three decimal places. If the win average is undefined, output "-". Output an empty line between cases.

Sample Input
2 4
1 rock 2 paper
1 scissors 2 paper
1 rock 2 rock
2 rock 1 scissors
2 1
1 rock 2 paper
0

Sample Output
0.333
0.667

0.000
1.000

#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
#include<queue>
#include<stack>
#include<algorithm>
#include<cmath>
#include<set>
#include<map>
#include<vector>
#include<fstream>
#include<sstream>
#include<iomanip>
#include <sstream>
using namespace std; const int N=; int player[N][]; // player[i][0]: win , player[i][1]: lose. int n,k;
short judge(string a, string b)
{
if(a == b)
{
return ;
}
short result;
if(a == "rock")
{
if(b == "scissors")
{
result = ;
}
else if(b == "paper")
{
result = -;
}
}
else if(a == "scissors")
{
if(b == "rock")
{
result = -;
}
else if(b == "paper")
{
result = ;
}
}
else
{
if(b == "scissors")
{
result = -;
}
else if(b == "rock")
{
result = ;
}
}
return result;
}
int main()
{
short ans;
string sa,sb;
int pa,pb;
bool first = true;
while(cin>>n && n)
{
if(!first)putchar('\n');
if(first)first=false;
memset(player,,sizeof(player));
cin>>k;
while(k--)
{
cin>>pa>>sa>>pb>>sb;
ans = judge(sa,sb);
if(ans)
{
if(ans==)//win
{
player[pa][]++;
player[pb][]++;
}
else//lose
{
player[pb][]++;
player[pa][]++;
}
}
}
for(int i=; i<=n; i++)
{
int total = player[i][]+player[i][];
if(total>)
{
printf("%.3f\n",((float)player[i][])/total);
}
else
{
printf("-\n");
}
}
}
return ;
}

Rock-Paper-Scissors Tournament[HDU1148]的更多相关文章

  1. 2018 ACM-ICPC 中国大学生程序设计竞赛线上赛 H题 Rock Paper Scissors Lizard Spock.(FFT字符串匹配)

    2018 ACM-ICPC 中国大学生程序设计竞赛线上赛:https://www.jisuanke.com/contest/1227 题目链接:https://nanti.jisuanke.com/t ...

  2. SDUT 3568 Rock Paper Scissors 状压统计

    就是改成把一个字符串改成三进制状压,然后分成前5位,后5位统计, 然后直接统计 f[i][j][k]代表,后5局状压为k的,前5局比和j状态比输了5局的有多少个人 复杂度是O(T*30000*25*m ...

  3. FFT(Rock Paper Scissors Gym - 101667H)

    题目链接:https://vjudge.net/problem/Gym-101667H 题目大意:首先给你两个字符串,R代表石头,P代表布,S代表剪刀,第一个字符串代表第一个人每一次出的类型,第二个字 ...

  4. Gym - 101667H - Rock Paper Scissors FFT 求区间相同个数

    Gym - 101667H:https://vjudge.net/problem/Gym-101667H 参考:https://blog.csdn.net/weixin_37517391/articl ...

  5. Gym101667 H. Rock Paper Scissors

    将第二个字符串改成能赢对方时对方的字符并倒序后,字符串匹配就是卷积的过程. 那么就枚举字符做三次卷积即可. #include <bits/stdc++.h> struct Complex ...

  6. 【题解】CF1426E Rock, Paper, Scissors

    题目戳我 \(\text{Solution:}\) 考虑第二问,赢的局数最小,即输和平的局数最多. 考虑网络流,\(1,2,3\)表示\(Alice\)选择的三种可能性,\(4,5,6\)同理. 它们 ...

  7. 题解 CF1426E - Rock, Paper, Scissors

    一眼题. 第一问很简单吧,就是每个 \(\tt Alice\) 能赢的都尽量让他赢. 第二问很简单吧,就是让 \(\tt Alice\) 输的或平局的尽量多,于是跑个网络最大流.\(1 - 3\) 的 ...

  8. HDOJ(HDU) 2164 Rock, Paper, or Scissors?

    Problem Description Rock, Paper, Scissors is a two player game, where each player simultaneously cho ...

  9. HDU 2164 Rock, Paper, or Scissors?

    http://acm.hdu.edu.cn/showproblem.php?pid=2164 Problem Description Rock, Paper, Scissors is a two pl ...

  10. 1090-Rock, Paper, Scissors

    描述 Rock, Paper, Scissors is a classic hand game for two people. Each participant holds out either a ...

随机推荐

  1. java1.8中Lambda表达式reduce聚合测试例子

    public class LambdaTest { public static void main(String[] args) { // 相当于foreach遍历操作结果值 Integer out ...

  2. uitableviewcell cell.accessoryType 右箭头

    实现右侧的小灰色箭头  只要将cell的accessoryType属性设置为 UITableViewCellAccessoryDisclosureIndicator就可以了. 代码为:cell.acc ...

  3. poj1308(简单并查集)

    题目链接:http://poj.org/problem?id=1308 题意:x, y 表示x 与 y连接,给出一波这样的数据,问这组数据能否构成树,即不能形成回路,不能有多个根节点:要注意可以是空树 ...

  4. 警告 - no rule to process file 'WRP_CollectionView/README.md' of type net.daringfireball.markdown for architecture i386

    warning: no rule to process file '/Users/mac/Downloads/Demo/Self/WRP_CollectionView/WRP_CollectionVi ...

  5. PHP面向对象编程之深入理解方法重载与方法覆盖(多态)

    这篇文章主要介绍了PHP面向对象编程之深入理解方法重载与方法覆盖(多态)的相关资料,需要的朋友可以参考下: 什么是多态? 多态(Polymorphism)按字面的意思就是"多种状态" ...

  6. maven File encoding has not been set

    原pom.xml配置文件: <?xml version="1.0" encoding="UTF-8"?> <project xmlns=&qu ...

  7. .net socket 层面实现代理服务器

    socket 层面实现代理服务器 首先是简一个简单的socket客户端和服务器端的例子 建立连接 Socket client = new Socket(AddressFamily.InterNetwo ...

  8. POJ 1655 Balancing Act 树的重心

    Balancing Act   Description Consider a tree T with N (1 <= N <= 20,000) nodes numbered 1...N. ...

  9. Arduino101学习笔记(十二)—— 101定时器中断

    一.API 1.开定时器中断 //*********************************************************************************** ...

  10. kylin学习笔记

    阅读官网,学到哪就写到哪 1.需要先建立Model 2.kylin需要配置事实表,纬度表:可以自定义join.  我的用法和官方建议的不同,我是直接在hive中将所有的取join成一个单表,再根据单表 ...