[今日头条] 2017客户端工程师实习生笔试题

华电北风吹

天津大学认知计算与应用重点实验室

日期：2016-09-05

题目一：

很简单，按照题目要求做即可AC.

#include <iostream>
#include <string>
using namespace std;

int main()
{
string str;
cin >> str;
int N;
cin >> N;
int p, l;
while (N--)
{
cin >> p >> l;
string in = "";
for (int j = p + l - 1; j >= p; j--)
{
in += str[j];
}
str = str.insert(p + l, in.c_str());
}
cout << str << endl;
return 0;
}

题目二：

也很简单，主要考察点是情况分类，只要细心吧所有的情况都考虑到就能AC.

#include <iostream>
#include <math.h>
using namespace std;

int main()
{
int n, s, L;
cin >> n >> s >> L;
int k = (L + 1) / (s + 1);
if (k % 13 == 0)
k--;
int last = n%k;
if (last == 0)
cout << ceil(1.0*n / k) << endl;
else
if (last % 13 != 0)
cout << ceil(1.0*n / k) << endl;
else
if (n / k == 0)
cout << ceil(1.0*n / k) + 1 << endl;
else
if (k == last + 1)
cout << ceil(1.0*n / k) + 1 << endl;
else
cout << ceil(1.0*n / k) << endl;
return 0;
}

题目三：

按照题目要求来即可。中间利用KMP降低时间复杂度就能AC.

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;

vector<int> nextVector(const string& pattern)
{
int m = pattern.size();
vector<int> result(m);
for (int i = 1; i < m; i++)
{
int j = result[i - 1];
while (j&&pattern[i] != pattern[j])
j = result[j - 1];
result[i] = pattern[i] == pattern[j] ? j + 1 : 0;
}
return result;
}

int KMP(const string& str, const string& pattern)
{
int n = str.size(), m = pattern.size();
vector<int> arr = nextVector(pattern);
int count = 0;
int p = 0;
for (int i = 0; i < n; i++)
{
while (p>0 && pattern[p] != str[i])
p = arr[p - 1];
if (str[i] == pattern[p])
p++;
if (p == m)
count++;
}
return count;
}

int main()
{
int n, k;
cin >> n >> k;
vector<string> data(n);
vector<int> permut(n);
for (int i = 0; i < n; i++)
{
cin >> data[i];
permut[i] = i;
}
int count = 0;
do  {
string pattern = "";
for (int i = 0; i < n; i++)
{
pattern += data[permut[i]];
}
string str = pattern + pattern;
if (KMP(str, pattern) == k + 1)
{
count++;
}
} while (next_permutation(permut.begin(), permut.end()));
cout << count << endl;
return 0;
}

题目四：

很有技巧性，把题目转化为位运算和二进制运算就容易AC了。

#include <iostream>
#include <vector>
using namespace std;

int main()
{
long long x, k;
cin >> x >> k;
vector<long long> bit;
long long flag = 1;
for (int i = 1; i < 64; i++)
{
if ((x&flag) == 0)
bit.push_back(flag);
flag = flag << 1;
}
flag = 1;
long long y = 0;
for (int i = 1; i < 64; i++)
{
if ((k&flag) > 0)
y += bit[i - 1];
flag = flag << 1;
}
cout << y << endl;
return 0;
}