1018. 錘子剪刀布 (20)
大家應該都會玩“錘子剪刀布”的遊戲:兩人同時給出手勢,勝負規則如圖所示:
現給出兩人的交鋒記錄,請統計雙方的勝、平、負次數,並且給出雙方分別出什麼手勢的勝算最大。
輸入格式:
輸入第1行給出正整數N(<=105),即雙方交鋒的次數。隨後N行,每行給出一次交鋒的信息,即甲、乙雙方同時給出的的手勢。C代表“錘子”、J代表“剪刀”、B代表“布”,第1個字母代表甲方,第2個代表乙方,中間有1個空格。
輸出格式:
輸出第1、2行分別給出甲、乙的勝、平、負次數,數字間以1個空格分隔。第3行給出兩個字母,分別代表甲、乙獲勝次數最多的手勢,中間有1個空格。如果解不唯一,則輸出按字母序最小的解。
輸入樣例:10 C J J B C B B B B C C C C B J B B C J J輸出樣例:
5 3 2 2 3 5 B B
/*
@ author: GaominQuan
@ data: 2014 - 2 - 21
@ mail: [email protected]
*/
#include<iostream>
#include<vector>
using namespace std;
enum types{B,C,J};
enum result{win,loss,same};
result race(types t1, types t2){
result rt;
if(t1 == t2){
rt = same;
}else{
if( t1 == B && t2 == C || t1 == C && t2 == J || t1 == J && t2 == B){
rt = win;
}else{
rt = loss;
}
}
return rt;
}
struct person{
int becauseWin[3];
int winTime;
int lossTime;
int equalTime;
person(){
winTime = 0;
lossTime = 0;
equalTime = 0;
for(int i = 0; i<3; i++){
//Notice Here !!!!!!!!!!!!!!!!!!!!!!!!!!! i and types which you need to use???
becauseWin[i] = 0;
}
};
};
types changeToTypes(char t1){
switch (t1){
case 'C':return C;
case 'B':return B;
case 'J':return J;
}
}
char chooseMaxWin(person p){
int maxTime = 0;
int maxType = 0;
for(int i = 0; i<3; i++){
if(p.becauseWin[i] > maxTime){
maxTime = p.becauseWin[i];
maxType = i;
}
}
switch(maxType){
case 0:return 'B';
case 1:return 'C';
case 2:return 'J';
}
}
int main(){
person p1,p2;
person winer,losser;
int N = 0;
cin>>N;
for(int i = 0; i<N; i++){
char t1,t2;
cin>>t1>>t2;
types type1 = changeToTypes(t1);
types type2 = changeToTypes(t2);
switch(race(type1,type2)){
case same:
p1.equalTime++;
p2.equalTime++;
break;
case win:
p1.winTime++;
p2.lossTime++;
p1.becauseWin[type1]++;
break;
case loss:
p2.winTime++;
p1.lossTime++;
p2.becauseWin[type2]++;
break;
}
}
cout<<p1.winTime<<" "<<p1.equalTime<<" "<<p1.lossTime<<endl;
cout<<p2.winTime<<" "<<p2.equalTime<<" "<<p2.lossTime<<endl;
cout<<chooseMaxWin(p1)<<" "<<chooseMaxWin(p2)<<endl;
return 0;
}