拼多多 100,100,5,30
第一题 奇偶数优先级排序 AC 100
import os
import sys
def Fn(data:list, N):
data.sort(reverse=True)
a = [str(x) for x in data if x%2==0]
b = [str(x) for x in data if x%2!=0]
data = a+b print(",".join(data[0:N]))
def main():
line = sys.stdin.readline().strip()
line = line.split(';')
data = list(map(int, line[0].split(',')))
N = int(line[1])
Fn(data, N) if __name__=='__main__':
main()
第二题 扑克牌 AC 100
#pragma once
#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
using namespace std;
void FnCore(string S1, const string &S2,
int N,
string newString,
string round,
vector<string>&res)
{
if (newString == S2 && round.size() == N)
{
res.push_back(round);
return;
}
if (S1.size() < 1 || round.size() > N)
return;
FnCore(S1.substr(1), S2, N, newString, round + 'd', res);
FnCore(S1.substr(1), S2, N, S1[0]+newString, round + 'l', res);
FnCore(S1.substr(1), S2, N, newString+ S1[0], round + 'r', res);
}
void Fn(string S1, string S2)
{
vector<string>res;
FnCore(S1, S2, S1.size(), "", "", res);
if (res.size() > 0)
sort(res.begin(), res.end());
for (int i = 0;i < res.size();++i)
{
for (int j = 0;j<res[i].size();++j)
{
printf("%c ", res[i][j]);
}
cout << endl;
}
}
int main()
{
int S;
cin >> S;
string S1, S2;
for (int i = 0;i < S;++i)
{
cin >> S1;
cin >> S2;
cout << "{" << endl;
Fn(S1, S2);
cout << "}" << endl;
}
return 0;
} 第三题 掷骰子 AC 5 import os
import sys
def FnCore(X, ret, count, index, CurMax):
if index==len(X):
count[CurMax-1]+=1
return
for i in range(1, X[index]+1):
ret.append(i)
if i>CurMax:
CurMax=i
FnCore(X, ret, count, index+1,CurMax)
del ret[-1]
def Fn(X):
maxVal = max(X)
count = [0]*maxVal
ret = []
CurMax= 0
FnCore(X, ret, count, 0,CurMax)
Total = sum(count)
P = 0
for i in range(len(count)):
P+=(i+1)*count[i]/Total
P = round(float(P), 2)
print(P)
def main():
n = int(input())
X = list(map(int, sys.stdin.readline().strip().split(' ')))
Fn(X)
if __name__=='__main__':
main() 第四题 寻找矩阵中第K大的数 AC 30 #pragma once
#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
using namespace std;
int Fn(int N, int M, int K)
{
priority_queue<int, vector<int>, less<int>>data;
for(int i = N;i>=1;--i)
for (int j = M;j >= 1;--j)
{
data.push(i*j);
}
for (int r = 0;r < K-1;++r)
{
data.pop();
}
int ret = data.top();
return ret;
}
int main()
{
int N, M, K;
cin >> N >> M >> K;
int ret = Fn(N, M, K);
cout << ret;
return 0;
} 
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