20200426腾讯数据分析与研发实习生笔试题及参考代码

第一题

题目描述

代码及注释

【1-V1】每场战役一操作

import sys
a, b = map(int, sys.stdin.readline().strip().split())
res, blood = 0, 0
for i in range(a):
    c, d = map(int, sys.stdin.readline().strip().split())
    if d*b > c: # 当收益大于花费时选择进入这场战役否则放弃
        res += d  # 获得的收益
        # 计算花费
        if blood >= c:    # 血量够时，直接减去需消耗的血量
            blood -= c
        else:             # 血量不够时，先购买血量直接减去需消耗的血量
            temp = (c-blood)//b+1
            res -= temp
            blood = (blood + temp*b)-c
print(res)

【1-V2】优化：计算所有参加的战役的血量消耗，最后统一根据消耗统一购买血量

import sys
a, b = map(int, sys.stdin.readline().strip().split())
res, blood = 0, 0
for i in range(a):
    c, d = map(int, sys.stdin.readline().strip().split())
    if d*b > c:     # 当收益大于花费时选择进入这场战役否则放弃
        res += d    # 获得的收益
        blood += c  # 消耗的血量

temp, rem = blood // b, blood % b
if rem == 0:
    print(res - temp)
else:
    print(res - temp - 1)

第二题

题目描述

代码及注释

【2-V1】 根据积分的定义编程实现数值积分，结果没问题但程序超时，本题应手推定积分公式，直接套公式计算(具体见解法二)

import sys

def calcArea1(A: int, a: int, b: int, n: int) -> int:
    deltax = (b-a)/n
    res = 0
    while a<b:
        temp = 2*A*a
        temp1 = pow(temp, 0.5)*2
        res += temp1*deltax
        a += deltax
    return res

def calcArea2(A: int, B: int, C: int, a: int, b: int, n: int) -> int:
    deltax = (b-a)/n
    res = 0
    while a<b:
        temp = 2*A*a
        temp1 = pow(temp, 0.5)
        temp2 = B*a+C
        res += (temp1-temp2)*deltax
        a += deltax
    return res

num = int(input())
for i in range(num):
    a, b, c = map(int, sys.stdin.readline().strip().split())
    delta = 1/(b**2)-2*c/(a*b)
    if delta <= 0:
        print(0)
    else:
        temp1 = pow(delta, 0.5) * a
        temp2 = a/b
        y1, y2 = temp2-temp1, temp2+temp1
        x1, x2 = y1/b-c/b, y2/b-c/b
        res, res1, n = 8, -1, 1024
        while abs(res-res1)/abs(res) >= 10**-5:
            res = res1
            n = 2*n
            res1 = calcArea1(a, 0, x1, n)
        res1 = res
        res, res2, n = 8, -1, 1024
        while abs(res-res2)/abs(res) >= 10**-5:
            res = res2
            n = 2*n
            res2 = calcArea2(a, b, c, x1, x2, n)
        print(res1+res)

【2-V2】利用定积分公式求解，从y1到y2对y进行积分

import sys

num = int(input())
for i in range(num):
    a, b, c = map(int, sys.stdin.readline().strip().split())
    delta = 1/(b**2)-2*c/(a*b)
    if delta <= 0:
        print(0)
    else:
        temp1 = pow(delta, 0.5) * a
        temp2 = a/b
        y1, y2 = temp2-temp1, temp2+temp1
        constant1, constant2, constant3 = 1/(2*b), -c/b, -1/(6*a)
        res2 = constant1*y2**2 + constant2*y2 + constant3*y2**3
        res1 = constant1*y1**2 + constant2*y1 + constant3*y1**3
        print(res2-res1)

第三题

题目描述

代码及注释

思路：
全排列（n个房间，每个房间m种选择：m^n）
不冲突的情况（第一个房间选择一个数[m种选择]，之后每个房间只要不与前一个数重复即可[每个房间m-1种选择，n-1个房间]：m*(m-1)^(n-1)）
冲突的方案总数 = 全排列 - 不冲突情况 = m^n - m*(m-1)^(n-1) = m * ( m^(n-1) - (m-1)^(n-1) )
超时优化思路：
快速幂，m^n 直接快速幂取模，m(m-1)^(n-1) 就对m取模，(m-1)^(n-1)取模，再把俩模乘起来加减就行了;
python的pow()自带快速幂

import sys
for line in sys.stdin:
    m, n = map(int, line.strip().split())
    temp1 = pow(m, n) % 100003
    temp2 = m % 100003
    temp3 = pow(m-1, n-1) % 100003
    res = (temp1 - temp2 * temp3) % 100003
    print(res)

第四题

题目描述

代码及注释

【4-V1】暴力 测例通过30%

import sys
n, k = map(int, sys.stdin.readline().strip().split())
data = []
for i in range(n):
    temp = list(map(int, sys.stdin.readline().strip().split()))
    data.append(temp)

res = 0
if n == 1:
    print(res)
for i in range(n):
    for j in range(i+1,n):
        if data[i][0]+data[j][0] == data[i][1]+data[j][1]:
            flag = True
            temp = data[i][0]+data[j][0]
            for kk in range(2,k):
                if data[i][kk]+data[j][kk] != temp:
                    flag = False
                    break
            if flag:
                res += 1
print(res)

【4-V2】 利用哈希表优化

思路：
将aij+akj = ai0+ak0变形为aij-ai0 = -(akj-ak0)
对每组数据将 (aij-ai0) 构建Hash表

import sys
from collections import defaultdict

n, k = map(int, sys.stdin.readline().strip().split())
data = defaultdict(lambda: 0)
tempB = [0]*(k-1)
res = 0
for i in range(n):
    tempA = list(map(int, sys.stdin.readline().strip().split()))
    for j in range(1, k):
        tempA[j] -= tempA[0]
        tempB[j-1] = -tempA[j]
    if tuple(tempB) in data:
        res += data[tuple(tempB)]
    data[tuple(tempA[1:])] += 1

print(res)

第五题

题目描述

代码及注释

from typing import List
import sys

def bfs(g: List[List[int]], g1: List[List[int]], visit: List[int], pos: int) -> int:
    res = 0
    table = []
    for i in g[pos]:
        if visit[i] == 0:
            visit[i] = 1
            res += 1
            table.append(i)
    for i in g1[pos]:
        if visit[i] == 0:
            visit[i] = 1
            res += 1
            table.append(i)
    for p in table:
        res += bfs(g, g1, visit, p)
    return res

t = int(input())
for i in range(t):
    n = int(input())
    data = []
    for j in range(n):
        x, y = map(int, sys.stdin.readline().strip().split())
        data.append([x, y])
    l = []
    d = {}
    for line in data:           # 获取所有用户的标号
        l.append(line[0])
        l.append(line[1])
    st = list(set(l))

    for i in range(len(st)):    # 将所有用户重新编号，按顺序编号
        d[st[i]] = i
    for i in range(len(data)):
        data[i][0] = d[data[i][0]]
        data[i][1] = d[data[i][1]]

    size = len(st)              # 构建无向图的邻接表
    g = [[] for i in range(size)]
    g1 = [[] for i in range(size)]
    for line in data:
        g[line[0]].append(line[1])
        g1[line[1]].append(line[0])

    res, tempRes = 0, 0         # 对无向图进行BFS搜索每个连通分量(朋友圈),记录最大的朋友圈
    visit = [0]*size
    for i in range(size):
        if visit[i] == 0:
            visit[i] = 1
            tempRes = 1 + bfs(g, g1, visit, i)
            res = max(res, tempRes)
    print(res)