PAT乙級題解(1018. 錘子剪刀布)
阿新 • • 發佈:2019-01-28
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資訊,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。
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這題賊煩(服氣臉)
#include <iostream> using namespace std; #define C 0 #define J 1 #define B 2 int wintimea[3] = { 0 }; int wintimeb[3] = { 0 }; int judge(char a, char b)//判決函式 { int judge; if (a == 'C' && b == 'B') { judge = -1;//b用B贏了 wintimeb[B]++; } else if (a == 'B' && b == 'J') { judge = -1;//b用J贏了 wintimeb[J]++; } else if (a == 'J' && b == 'C') { judge = -1;//b用C贏了 wintimeb[C]++; } else if (a == 'B' && b == 'C') { judge = 1;//a用B贏了 wintimea[B]++; } else if (a == 'J' && b == 'B') { judge = 1;//a用J贏了 wintimea[J]++; } else if (a == 'C' && b == 'J') { judge = 1;//a用C贏了 wintimea[C]++; } else if (a == b) { judge = 0; } return judge; } int main() { int n; int result; cin >> n; char a, b; int wina = 0, winb = 0, draw = 0; for (int i = 0; i < n; i++) { cin >> a >> b; result = judge(a, b); if (result == 0) draw++; else if (result < 0) winb++; else wina++; } cout << wina << " " << draw << " " << winb << endl; cout << winb << " " << draw << " " << wina << endl; char MaxA = ((wintimea[C]>wintimea[B]&&wintimea[C]>=wintimea[J])?'C':(wintimea[B]>=wintimea[J]?'B':'J')); char MaxB = ((wintimeb[C]>wintimeb[B]&&wintimeb[C]>=wintimeb[J])?'C':(wintimeb[B]>=wintimeb[J]?'B':'J')); cout << MaxA << " " << MaxB; return 0; }