一、问题定义
大家应该都会玩“锤子剪刀布”的游戏:两人同时给出手势,胜负规则如下:
剪刀 > 布, 锤子 > 剪刀, 布 > 锤子
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。
输入格式:
输入第 1 行给出正整数 N(≤),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。
输出格式:
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。
输入样例:
10
C J
J B
C B
B B
B C
C C
C B
J B
B C
J J
输出样例:
5 3 2
2 3 5
B B
二、解题
python:最后一例超时
def jCB(a, b):
if a == b:
return states[2]
else:
return Table[(a, b)]
def sucessGeture(suc):
max_cnt = max(list(suc.values()))
max_gess = [key for key in suc.keys() if suc[key] == max_cnt]
return sorted(max_gess)[0]
def myPrint(array):
print("{} {} {}".format(array[0], array[1], array[2]))
cnt = input()
Alpha = [0, 0, 0] # 甲乙
Beta = [0, 0, 0]
Suc_Alpha = {ges:0 for ges in ["J", "C", "B"]} # count the succeed hand
Suc_Beta = {ges:0 for ges in ["J", "C", "B"]}
states = [(1, -1), (-1, 1), (0, 0)] # >, <, =
Table = { ("C", "J"):states[0], ("C", "B"):states[1],
("J", "C"):states[1], ("B", "C"):states[0],
("B", "J"):states[1], ("J", "B"):states[0]}
for j in range(int(cnt)):
Gestures = input()
A, B = Gestures[0], Gestures[2]
Ans_a, Ans_b = jCB(A, B)[0], jCB(A, B)[1] # 1 0 -1
if Ans_a == 1:
Alpha[0] += 1 # succeed
Beta[2] += 1
Suc_Alpha[A] += 1 # count success gesture
elif Ans_b == 1:
Beta[0] += 1 # failed
Alpha[2] += 1
Suc_Beta[B] += 1
elif Ans_a == 0: # equal
Alpha[1] += 1
Beta[1] += 1
myPrint(Alpha)
myPrint(Beta)
print(sucessGeture(Suc_Alpha), sucessGeture(Suc_Beta))
C++
#include <iostream>
using namespace std;
int count(char *JCB, char hand_chr)
{
int length = 3;
for ( int i = 0; i < length; i++) {
if (JCB[i] == hand_chr) {
return i;
}
}
}
int findMax(int *suc)
{ // linear search
int max = suc[0];
for (int i = 1; i < 3; i++) {
if (max < suc[i]) {
max = suc[i];
}
}
return max;
}
char findMinChr(char *max_chrs, int max_cnt)
{ // find min_chr in max_chrs
if (max_cnt == 0) {
return 'B';
} else {
char min_chr = max_chrs[0];
for (int i = 1; i < max_cnt ; i++) {
if (min_chr > max_chrs[i]) {
min_chr = max_chrs[i];
}
}
return min_chr;
}
}
int main(int argc, char *argv[])
{
int N, max_A, max_B, max_cnt= 0;
char max_chrs[3];
char max_chr_A, max_chr_B = ' ';
int cnt_A[3] = {0, 0, 0};
int cnt_B[3] = {0, 0, 0};
int suc_A[3] = {0, 0, 0};
int suc_B[3] = {0, 0, 0};
char JCB[3] = {'J', 'C', 'B'};
char A, B = ' ';
cin >> N; //cout << N << "
";
// C J B
for (int i = 0; i < N; i++) {
cin >> A >> B; //cout << A << " " << B << "
";
if (A == B) {
cnt_A[1] += 1;
cnt_B[1] += 1;
} else if ((A == 'C' && B == 'J') ||
(A == 'J' && B == 'B') ||
(A == 'B' && B == 'C')) {
cnt_A[0] += 1;
cnt_B[2] += 1;
suc_A[count(JCB, A)] += 1; // succeed count
} else if ((A == 'C' && B == 'B') ||
(A == 'J' && B == 'C') ||
(A == 'B' && B == 'J')) {
cnt_A[2] += 1;
cnt_B[0] += 1;
suc_B[count(JCB, B)] += 1;
}
}
cout << cnt_A[0] << " " << cnt_A[1] << " " << cnt_A[2] << "
";
cout << cnt_B[0] << " " << cnt_B[1] << " " << cnt_B[2] << "
";
max_A = findMax(suc_A);
max_B = findMax(suc_B);
// 有重复代码
for (int i, j = 0; i < 3; i++) {
if (suc_A[i] == max_A) {
max_chrs[j] = JCB[i];
j++;
max_cnt++;
}
}
max_chr_A = findMinChr(max_chrs, max_cnt);
max_cnt = 0;
for (int i, j = 0; i < 3; i++) {
if (suc_B[i] == max_B) {
max_chrs[j] = JCB[i];
j++;
max_cnt++;
}
}
max_chr_B = findMinChr(max_chrs, max_cnt);
cout << max_chr_A << " " << max_chr_B;
return 0;
}