ZOJ1111:Poker Hands(模拟题)
A poker deck contains 52 cards - each card has a suit which is one of clubs, diamonds, hearts, or spades (denoted C, D, H, S in the input data). Each card also has a value which is one of 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king, ace (denoted 2, 3, 4, 5, 6, 7, 8, 9, T, J, Q, K, A). For scoring purposes, the suits are unordered while the values are ordered as given above, with 2 being the lowest and ace the highest value.
A poker hand consists of 5 cards dealt from the deck. Poker hands are ranked by the following partial order from lowest to highest
- High Card. Hands which do not fit any higher category are ranked by the value of their highest card. If the highest cards have the same value, the hands are ranked by the next highest, and so on.
- Pair. 2 of the 5 cards in the hand have the same value. Hands which both contain a pair are ranked by the value of the cards forming the pair. If these values are the same, the hands are ranked by the values of the cards not forming the pair, in decreasing order.
- Two Pairs. The hand contains 2 different pairs. Hands which both contain 2 pairs are ranked by the value of their highest pair. Hands with the same highest pair are ranked by the value of their other pair. If these values are the same the hands are ranked by the value of the remaining card.
- Three of a Kind. Three of the cards in the hand have the same value. Hands which both contain three of a kind are ranked by the value of the 3 cards.
- Straight. Hand contains 5 cards with consecutive values. Hands which both contain a straight are ranked by their highest card.
- Flush. Hand contains 5 cards of the same suit. Hands which are both flushes are ranked using the rules for High Card.
- Full House. 3 cards of the same value, with the remaining 2 cards forming a pair. Ranked by the value of the 3 cards.
- Four of a kind. 4 cards with the same value. Ranked by the value of the 4 cards.
- Straight flush. 5 cards of the same suit with consecutive values. Ranked by the highest card in the hand.
Your job is to compare several pairs of poker hands and to indicate which, if either, has a higher rank.
Input Specification
Several lines, each containing the designation of 10 cards: the first 5 cards are the hand for the player named "Black" and the next 5 cards are the hand for the player named "White."
Output Specification
For each line of input, print a line containing one of:
Black wins.
White wins.
Tie.
Sample Input
2H 3D 5S 9C KD 2C 3H 4S 8C AH
2H 4S 4C 2D 4H 2S 8S AS QS 3S
2H 3D 5S 9C KD 2C 3H 4S 8C KH
2H 3D 5S 9C KD 2D 3H 5C 9S KH
Sample Output
White wins.
Black wins.
Black wins.
Tie.
题意:又是一道模拟题,好久没敲过模拟题了,这次敲了蛮久,还记得暑假由此做某个地方的省赛,也是一道模拟题,那道题是打麻将,这次变成玩扑克了。
题意很简单,看过赌神的人都知道,每人手中5张排,比牌面大小,牌面由大到小分别是(这里花色无大小)
同花顺:牌面一样,只比较最大的看谁大,一样大则平手
四条:四个一样的,看这四个一样的中谁大谁赢
葫芦:三条+一对,看三条中谁的牌面大
同花:都是同花就按从大到小比较,看谁的更大
顺子:都是顺子看最大的谁大,一样则平手
三条:三条看三条中谁的牌面大
两对:两对就看谁的对子大,都一样大比单牌
一对:比对子,一样则比单牌
单排:按顺序比较大小,大者胜
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std; struct node
{
int num,mark;
} white[10],black[10]; int cmp(node a,node b)
{
return a.num<b.num;
} int set(char c)
{
if(c<='9' && c>='2')
return c-'0';
if(c == 'T')
return 10;
if(c == 'J')
return 11;
if(c == 'Q')
return 12;
if(c == 'K')
return 13;
if(c == 'A')
return 14;
if(c == 'C')
return 1;
if(c == 'D')
return 2;
if(c == 'H')
return 3;
if(c == 'S')
return 4;
} int up[10]; int same(node *a)
{
int i,l = 0;
for(i = 2; i<=5; i++)
if(a[i].num!=a[i-1].num)
up[l++] = i;
return l;
}
/*
我们用数字表示牌面
1-同花顺
2-四条
3-葫芦
4-同花
5-顺子
6-三条
7-两对
8-一对
9-单牌
*/
int judge(node *a)
{
int flag1 = 0;
int flag2 = 0;
if(a[1].mark == a[2].mark && a[2].mark == a[3].mark &&a[3].mark == a[4].mark &&a[4].mark == a[5].mark)
flag1 = 1;
if(a[1].num+1 == a[2].num && a[2].num+1 == a[3].num &&a[3].num+1 == a[4].num &&a[4].num+1 == a[5].num)
flag2 = 1;
if(flag1 && flag2)
return 1;
else if(flag1 && !flag2)
return 4;
else if(flag2 && !flag1)
return 5;
int k = same(a);
if(k == 0)
return 2;
else if(k == 1)
{
if(up[0] == 2 || up[0] == 5)
return 2;
if(up[0] == 3 || up[0] == 4)
return 3;
}
else if(k == 2)
{
if(up[0] == 2 && up[1] == 3 || up[0] == 4 && up[1] == 5 || up[0] == 2 && up[1] == 5)
return 6;
else
return 7;
}
else if(k == 3)
return 8;
else if(k == 4)
return 9;
return 0;
} int find_one(node *a)
{
int i;
for(i = 1; i<=5; i++)
{
if(i == 1 && a[i].num!=a[i+1].num)
return 1;
if(i == 5 && a[i].num!=a[i-1].num)
return 5;
if(a[i].num!=a[i-1].num && a[i].num!=a[i+1].num)
return i;
}
return 0;
} int find_pair(node *a)
{
int i;
for(i = 1; i<=5; i++)
{
if(i == 1 && a[i].num == a[i+1].num)
return 2;
if(a[i].num == a[i-1].num)
return i;
}
return 0;
} int compare(int x)
{
int i,j,a,b;
if(x == 1 || x == 5)
{
if(white[5].num>black[5].num)
return -1;
else if(white[5].num<black[5].num)
return 1;
return 0;
}
else if(x == 2 || x == 3 || x == 6)
{
if(white[3].num>black[3].num)
return -1;
else if(white[3].num<black[3].num)
return 1;
return 0;
}
else if(x == 4 || x == 9)
{
for(i = 5; i>=1; i--)
{
if(black[i].num<white[i].num)return -1;
else if(black[i].num>white[i].num)
return 1;
}
return 0;
}
else if(x == 7)
{
if(white[4].num>black[4].num)
return -1;
else if(white[4].num<black[4].num)
return 1;
else
{
if(white[2].num>black[2].num)
return -1;
else if(white[2].num<black[2].num)
return 1;
int a = find_one(white);
int b = find_one(black);
if(black[b].num<white[a].num)
return -1;
else if(black[b].num>white[a].num)
return 1;
return 0;
}
}
else if(x == 8)
{
int a = find_pair(white);
int b = find_pair(black);
int tem_w[10],tem_b[10],i,lb = 1,lw = 1;
if(white[a].num>black[b].num)
return -1;
else if(white[a].num<black[b].num)
return 1;
for(i = 1; i<=5; i++)
{
if(i!=a && i!=a-1)
tem_w[lw++] = white[i].num;
if(i!=b && i!=b-1)
tem_b[lb++] = black[i].num;
}
for(i = 3; i>=1; i--)
{
if(tem_w[i]>tem_b[i])
return -1;
else if(tem_b[i]>tem_w[i])
return 1;
}
return 0;
}
return 0;
} int main()
{
char str[5];
int sumb,sumw,j,k,i;
while(~scanf("%2s",str))
{
j = 1;
k = 1;
black[j].num = set(str[0]);
black[j++].mark = set(str[1]);
for(i = 2; i<=10; i++)
{
scanf("%2s",str);
if(i<6)
{
black[j].num = set(str[0]);
black[j++].mark = set(str[1]);
}
else
{
white[k].num = set(str[0]);
white[k++].mark = set(str[1]);
}
}
sort(black+1,black+6,cmp);
sort(white+1,white+6,cmp);
sumb = judge(black);
sumw = judge(white);
if(sumb>sumw)
printf("White wins.\n");
else if(sumb<sumw)
printf("Black wins.\n");
else
{
int flag = compare(sumb);
if(flag == -1)
printf("White wins.\n");
else if(flag == 1)
printf("Black wins.\n");
else
printf("Tie.\n");
}
} return 0;
}
ZOJ1111:Poker Hands(模拟题)的更多相关文章
- poj 1008:Maya Calendar(模拟题,玛雅日历转换)
Maya Calendar Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 64795 Accepted: 19978 D ...
- poj 1888 Crossword Answers 模拟题
Crossword Answers Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 869 Accepted: 405 D ...
- CodeForces - 427B (模拟题)
Prison Transfer Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u Sub ...
- sdut 2162:The Android University ACM Team Selection Contest(第二届山东省省赛原题,模拟题)
The Android University ACM Team Selection Contest Time Limit: 1000ms Memory limit: 65536K 有疑问?点这里 ...
- 全国信息学奥林匹克联赛 ( NOIP2014) 复赛 模拟题 Day1 长乐一中
题目名称 正确答案 序列问题 长途旅行 英文名称 answer sequence travel 输入文件名 answer.in sequence.in travel.in 输出文件名 answer. ...
- UVALive 4222 Dance 模拟题
Dance 题目连接: https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&pag ...
- cdoj 25 点球大战(penalty) 模拟题
点球大战(penalty) Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.uestc.edu.cn/#/problem/show/2 ...
- Educational Codeforces Round 2 A. Extract Numbers 模拟题
A. Extract Numbers Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/600/pr ...
- URAL 2046 A - The First Day at School 模拟题
A - The First Day at SchoolTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudg ...
随机推荐
- shorter concat [reverse longer]
shorter concat [reverse longer] Description: Given 2 strings, a and b, return a string of the form: ...
- 设置一个POJO的某个属性的默认值
//月利率private BigDecimal monthRate=new BigDecimal(0);
- 函数hash_get_nth_cell
/************************************************************//** Gets the nth cell in a hash table. ...
- poj3666
一道不错的dp题 就是最小修改代价,使序列变为一个非下降序或非上升(由于数据较弱直接求非下降即可,当然非上升非下降本质是一样的) 观察可得到,修改后得到的数列中的元素最后一定都在原序列中: 由此我们可 ...
- ASP.NET的六种验证控件的使用
C# 中的验证控件分为一下六种 :1 CompareValidator:比较验证,两个字段的值是否相等,比如判断用户输入的密码和确认密码是否一致,则可以用改控件: 2 CustomValidator ...
- 微软 Virtual studion Code
在 Build 2015 大会上,微软除了发布了 Microsoft Edge 浏览器和新的 Windows 10 预览版外,最大的惊喜莫过于宣布推出免费跨平台的 Visual Studio Code ...
- Java [leetcode 17]Letter Combinations of a Phone Number
题目描述: Given a digit string, return all possible letter combinations that the number could represent. ...
- codeforces 672C - Recycling Bottles 贪心水题
感觉很简单,就是讨论一下 #include <stdio.h> #include <string.h> #include <algorithm> #include ...
- HDU5046 Airport dancing links 重复覆盖+二分
这一道题和HDU2295是一样 是一个dancing links重复覆盖解决最小支配集的问题 在给定长度下求一个最小支配集,只要小于k就行 然后就是二分答案,每次求最小支配集 只不过HDU2295是浮 ...
- 不区分大小写匹配字符串,并在不改变被匹配字符串的前提下添加html标签
问题描述:最近在搭建一个开源平台网站,在做一个简单搜索的功能,需要将搜索到的结果中被匹配的字符串添加不一样的颜色,但是又不破坏被匹配的字符串. 使用的方法是替换被匹配的字符串加上font标签.但是搜索 ...