360笔试0822
第一题是合法字符串问题，直接暴力遍历判断就可以了
#include <iostream>
#include<string>
using namespace std;
int main()
{
int n;
cin >> n;
int cnt = 0;
for (int i = 0; i < n; i++)
{
string str;
cin >> str;
if (str.size() >= 1 && str.size() <= 10)
{
int sz = str.size();
int j = 0;
for (; j < sz; j++)
{
int ch = (int)str[j];
if ((ch >= 'A' && ch <= 'Z') || (ch >= 'a' && ch <= 'z'))
{
continue;
}
else
{
break;
}
}
if (j == sz)
{
cnt++;
}
}
}
cout << cnt;
return 0;
}
第二题暴力的话能A 55%，后面优化了一下，主要是消除重复的无效操作，但不知道有没有提交成功
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
int n, m;
cin >> n >> m;
vector<int> nums(n);
for (int i = 1; i <= n; i++)
{
nums[i - 1] = i;
}
vector<int> datas;
for (int i = 0; i < m; i++)
{
int tmp;
cin >> tmp;
datas.push_back(tmp);
}
vector<int> tasks;
int cnt1 = 0;
int cnt2 = 0;
int last = -1;
for (int i = 0; i < m; i++)
{
int now = datas[i];
if (now == last && last == 2)
{
cnt2++;
if (i == m - 1)
{
if (cnt2 % 2)
{
tasks.push_back(2);
}
}
}
else if (now == last && last == 1)
{
cnt1++;
if (i == m - 1)
{
cnt1 %= n;
for (int j = 0; j < cnt1; j++)
{
tasks.push_back(1);
}
}
}
else if (now != last)
{
if (now == 1)
{
if (cnt2 % 2)
{
tasks.push_back(2);
}
cnt2 = 0;
cnt1 = 1;
if (i == m - 1)
{
tasks.push_back(1);
}
}
else
{
cnt1 %= n;
for (int j = 0; j < cnt1; j++)
{
tasks.push_back(1);
}
cnt2 = 1;
cnt1 = 0;
if (i == m - 1)
{
tasks.push_back(2);
}
}
}
last = now;
}
int sz = tasks.size();
for (int i = 0; i < sz; i++)
{
int tmp = tasks[i];
if (tmp == 1)
{
int first = nums[0];
for (int i = 1; i < n; i++)
{
nums[i - 1] = nums[i];
}
nums[n - 1] = first;
}
else
{
for (int j = 0; j < n; j += 2)
{
swap(nums[j], nums[j + 1]);
}
}
}
for (int i = 0; i < n; i++)
{
cout << nums[i] << " ";
}
return 0;
}
查看11道真题和解析
