腾讯计算机视觉笔试题目,笔试-腾讯-180916 - 算法/深度学习/NLP面试笔记

不定项 25，编程 3

字符串系数

暴力 KMP(70%)

def get_nxt(T):

n = len(T)

nxt = [0] * n

max_len = 0

for i in range(1, n):

while max_len > 0 and T[max_len] != T[i]:

max_len = nxt[max_len - 1]

if T[i] == T[max_len]:

max_len += 1

nxt[i] = max_len

return nxt

def kmp(S, T):

pos = []

nxt = get_nxt(T)

cnt = 0

for i in range(len(S)):

while cnt > 0 and T[cnt] != S[i]:

cnt = nxt[cnt - 1]

if T[cnt] == S[i]:

cnt += 1

if cnt == len(T):

pos.append(i - len(T) + 1)

cnt = nxt[cnt - 1]

return pos

k = int(input())

A = input()

B = input()

res = dict()

for i in range(len(A) - k + 1):

p = A[i: i + k]

# print(p)

if p not in res:

res[p] = len(kmp(B, p))

print(sum(res.values()))

暴力-使用C++库函数(AC)

size_t str_count(const string& S, const string& T) {

size_t cnt = 0;

for (size_t i = 0; (i = S.find(T, i)) != string::npos; i++, cnt++);

return cnt;

}

int main() {

int k = 2;

//cin >> k;

string A{“abab”};

//cin >> A;

string B{“ababab”};

//cin >> B;

vector tmp;

for (int i = 0; i < A.length() - k + 1; i++)

tmp.push_back(A.substr(i, k));

cout << tmp.size() << endl;

unique(tmp.begin(), tmp.end()); // 去重，这里直接使用 set 应该也可以

size_t ans = 0;

for (auto T : tmp) {

ans += str_count(A, T);

}

cout << ans << endl;

//system("PAUSE");

return 0;

}

## 小Q与牛牛的游戏

**Python**(AC)

```python

def solve(x, y):

""""""

ts = int(((x + y) * 2) ** 0.5)

if (ts * (ts + 1)) != (x + y) * 2:

return -1

cnt = 0

if x < ts:

return 1

while x > 0:

x -= ts

ts -= 1

cnt += 1

return cnt

x, y = list(map(int, input().split()))

print(solve(x, y))

三元组

Python(50%)

def solve(x, y, z):

""""""

cnt = 0

for i in range(1, x + 1):

for j in range(1, y + 1):

m_min = abs(i - j) + 1

m_max = min(i+j-1, z)

cnt = (cnt + m_max - m_min + 1) % 1000000007

return cnt

x, y, z = sorted(list(map(int, input().split())))

print(solve(x, y, z))

C++(60%)

int solve(int x, int y, int z)

{

int cnt = 0;

for (int i = 1; i <= x; ++i) {

for (int j = 1; j <= y; ++j) {

int m_min = abs(i - j) + 1;

int m_max = min(i + j - 1, z);

cnt = (cnt + m_max - m_min + 1) % 1000000007;

}

}

return cnt;

}

int main()

{

vector xyz;

int times = 3;

while (times--) {

int num;

cin >> num;

xyz.push_back(num);

}

sort(xyz.begin(), xyz.end());

int x = xyz[0];

int y = xyz[1];

int z = xyz[2];

cout << solve(x, y, z) << endl;

return 0;

}