1018 锤子剪刀布 (20分)
标签：nts 注释 ack ensure oid 重载 命名 ali highlight
大家应该都会玩“锤子剪刀布”的游戏：两人同时给出手势，胜负规则如图所示：
现给出两人的交锋记录，请统计双方的胜、平、负次数，并且给出双方分别出什么手势的胜算最大。
输入第 1 行给出正整数 N（≤10?5??），即双方交锋的次数。随后 N 行，每行给出一次交锋的信息，即甲、乙双方同时给出的的手势。C 代表“锤子”、J 代表“剪刀”、B 代表“布”，第 1 个字母代表甲方，第 2 个代表乙方，中间有 1 个空格。
输出第 1、2 行分别给出甲、乙的胜、平、负次数，数字间以 1 个空格分隔。第 3 行给出两个字母，分别代表甲、乙获胜次数最多的手势，中间有 1 个空格。如果解不唯一，则输出按字母序最小的解。
这题很讨厌，看着简单，但是要判断很多种情况，这次在写的时候虽然小心了点，但是没有边写边调试各个功能，导致之后多调半个小时，差点以为这题要流掉了
代码部分都给出了注释，因为能力不够，代码长了点，见谅
//甲方命名theone，乙方命名：theshy
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
//C:锤子 B:布 J：剪刀
bool compare(char a,char b[])
{
for (int i = 0; i < 3; i++)
{
if (a == b[i])
{
return true;
}
}
return false;
}
//求取赢取最多次数的手势所用
int max_three(int C, int B, int J)//fun=0,返回C。fun=1,返回B。fun=2,返回J。
{
if (C > B&&C>J||C==J&&C>B)
{
return 0;
}
else if (B > C&& B > J||B==C&&B>J||B==J&&B>C)
{
return 1;
}
else if (J > C&& J > B)
{
return 2;
}
}
//当甲方出拳头的情况进行判断
void judge_C(char b, int& win_o, int& win_s, int& lose_o, int& lose_s, int& equal_o, int& equal_s,int& count_o_C,int& count_s_B)
{
if (b == ‘C‘)
{
equal_o++;
equal_s++;
}
else if ( b == ‘B‘)
{
lose_o++;
win_s++;
count_s_B++;
}
else if (b == ‘J‘)
{
win_o++;
lose_s++;
count_o_C++;
}
}
//当甲方出布的情况
void judge_B( char b, int& win_o, int& win_s, int& lose_o, int& lose_s, int& equal_o, int& equal_s,int& count_o_B,int& count_s_J)
{
if (b == ‘B‘)
{
equal_o++;
equal_s++;
}
else if (b == ‘C‘)
{
win_o++;
lose_s++;
count_o_B++;
}
else if (b == ‘J‘)
{
lose_o++;
win_s++;
count_s_J++;
}
}
//当甲方出剪刀的情况
void judge_J( char b, int& win_o, int& win_s, int& lose_o, int& lose_s, int& equal_o, int& equal_s,int& count_o_J,int& count_s_C)
{
if (b == ‘J‘)
{
equal_o++;
equal_s++;
}
else if (b == ‘B‘)
{
win_o++;
lose_s++;
count_o_J++;
}
else if (b == ‘C‘)
{
lose_o++;
win_s++;
count_s_C++;
}
}
//获取theone所赢最多的手势
void get_gesture_o(int a,int b,int c,char &gesture_o)//a ,b,c分别代表C,B,J各手势所赢的次数，gesture_o保存所赢手势
{
if (max_three(a, b, c) == 0) //1表示C：锤子, 0表示B：布， 2表示J：剪刀
{
gesture_o = ‘C‘;
}
else if (max_three(a, b, c) == 1)
{
gesture_o = ‘B‘;
}
else if (max_three(a, b, c) == 2)
{
gesture_o = ‘J‘;
}
}
//获取theshy所赢最多的手势
void get_gesture_s(int a, int b, int c, char& gesture_s)//a ,b,c分别代表C,B,J各手势所赢的次数，gesture_o保存所赢手势
{
if (max_three(a, b, c) == 0) //0表示C：锤子, 1表示B：布， 2表示J：剪刀
{
gesture_s = ‘C‘;
}
else if (max_three(a, b, c) == 1)
{
gesture_s = ‘B‘;
}
else if (max_three(a, b, c) == 2)
{
gesture_s = ‘J‘;
}
}
int judger(vector<char> &one,vector<char> ­, int& win_o, int& win_s, int& lose_o, int& lose_s, int& equal_o, int& equal_s ,char &gesture_o,char &gesture_s,char data[])
{
int count_o_C = 0, count_o_B = 0, count_o_J = 0;
int count_s_C = 0, count_s_B = 0, count_s_J = 0;
for (int i = 0; i < one.size(); i++)
{
//判断是否是所要求C,B,J其中一个
if (!compare(one[i], data) || !compare(shy[i], data))
{
return -1;
}
//获取赢，平，输的次数
else
{
if (one[i] == ‘C‘)
{
judge_C(shy[i],win_o,win_s,lose_o,lose_s,equal_o,equal_s,count_o_C,count_s_B);
}
else if (one[i] == ‘B‘)
{
judge_B(shy[i], win_o, win_s, lose_o, lose_s, equal_o, equal_s,count_o_B,count_s_J);
}
else if(one[i]==‘J‘)
{
judge_J(shy[i], win_o, win_s, lose_o, lose_s, equal_o, equal_s,count_o_J,count_s_C);
}
}
}
//获取手势
get_gesture_o(count_o_C, count_o_B, count_o_J, gesture_o);
get_gesture_s(count_s_C, count_s_B, count_s_J, gesture_s);
/*cout << "count_o_C:" << count_o_C << " " << "count_o_B:" << count_o_B << " " << "count_o_J:" << count_o_J << endl;
cout << "count_s_C:" << count_s_C << " " << "count_s_B:" << count_s_B << " " << "count_s_J:" << count_s_J << endl;*/
return 0;
}
int main()
{
vector<char> theone, theshy;//甲方，乙方
char a = 0, b = 0;//用作对vector容器的输入中介，vector容器用下标表示的时候不能进行输入操作，这是在标准库中定义，如果你技术够可以自己写一个重载函数
int N = 0;//比较次数
cin >> N;
int win_o = 0, win_s = 0, lose_o = 0, lose_s = 0, equal_o = 0, equal_s = 0;//甲方所赢次数，甲方所输次数，甲方平局次数，乙方所赢次数，乙方所输次数，乙方平局次数
char gesture_o = 0, gesture_s = 0;//甲方赢得最多的手势，乙方...
char data[3] = { ‘B‘,‘J‘,‘C‘ };//判断输入是否满足题意所用
//输入
for (int i = 0; i < N; i++)
{
cin >> a >> b;
theone.push_back(a);
theshy.push_back(b);
}
int result = 0;
//函数调用
result = judger(theone, theshy, win_o, win_s, lose_o, lose_s, equal_o, equal_s, gesture_o, gesture_s, data);//判断输入是否是C，B,J之中
//判断返回值
if (!result)//观察函数的返回值，返回值为0的时候表示函数正确调用
{
//输出
cout << win_o << " " << equal_o << " " << lose_o << endl;
cout << win_s << " " << equal_s << " " << lose_s << endl;
cout << gesture_o << " " << gesture_s;
}
else
{
return 0;
}
return 0;
}
标签：nts 注释 ack ensure oid 重载 命名 ali highlight
原文地址：https://www.cnblogs.com/zongji/p/12243730.html
