题目:
输入样例:
[╮][╭][o][~\][/~] []
[╯][╰][^][-][=][>][<][@][⊙]
[Д][▽][_][ε][^] ...
4
1 1 2 2 2
6 8 1 5 5
3 3 4 3 3
2 10 3 9 3
输出样例:
╮(╯▽╰)╭
<(@Д=)/~
o(^ε^)o
Are you kidding me? @\/@
Python
先用正则匹配到各部位表情,然后尝试输出完整表情,出现异常就是数组下标越界了,此时输出Are you kidding me? @\/@
import re
pat = re.compile(r'\[(.+?)\]')
hands = pat.findall(input())
eyes = pat.findall(input())
mouths = pat.findall(input())
for i in range(int(input())):
a, b, c, d, e = [int(x) - 1 for x in input().split()]
try:
print(hands[a] + '(' + eyes[b] + mouths[c] + eyes[d] + ')' + hands[e])
except IndexError:
print(r'Are you kidding me? @\/@')
emmm,提交上去三个测试点全是非零返回,你在逗我吗?@\/@
参考了玮智能老哥的博客,得知原因可能是样例输入含有非ASCII字符,而PAT服务器的python解释器无法正常读入这些字符,详情点击这里
C++
哼(╯‵□′)╯︵┻━┻,python不行我还不能用c++写吗,用的柳婼小姐姐的方法,创键一个装vector的vector容器,这样数组的size是可变的,然后就是遍历各行表情,将各部位表情装入vector,然后判断数组下表是否越界,分别处理就行了
#include
#include
using namespace std;
int main() {
vector<vector> v;
for (int i = 0; i < 3; i++) {
string s;
getline(cin, s);
vector row;
int j = -1, k = 0;
while (++j < s.length()) {
if (s[j] == '[') {
while (++k > n;
for (int i = 0; i > a >> b >> c >> d >> e;
if (a > v[0].size() || b > v[1].size() || c > v[2].size() || d > v[1].size() || e > v[0].size() || a < 1 || b < 1 || c < 1 || d < 1 || e < 1)
cout << "Are you kidding me? @\\/@" << endl;
else
cout << v[0][a-1] << '(' << v[1][b-1] << v[2][c-1] << v[1][d-1] << ')' << v[0][e-1] << endl;
}
return 0;
}