备战蓝桥杯大赛：Python必刷题单之基础算法
枚举
百钱买百鸡
题目链接
# 枚举公鸡 21次
for x in range(0, 21):
# 枚举母鸡 34
for y in range(0, 34):
# 枚举小鸡
z = 100 - x - y
# 找到符合条件的解
if 5 * x + 3 * y + z // 3 == 100 and z % 3 == 0:
print(x, y, z)
数组查找及替换
题目链接
n, b = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
for x in a:
if x % b == 0:
continue
if x >= ord('A') and x <= ord('Z'):
print(chr(x), end = " ")
else:
print(x, end = " ")
校门外的树
题目链接
n, m = map(int, input().split())
a = [0] * (n + 1)
for i in range(m):
L, R = map(int, input().split())
for j in range(L, R + 1):
a[j] = 1
ans = 0
for i in range(n + 1):
if a[i] == 0:
ans += 1
print(ans)
铺地毯
题目链接
n = int(input())
a = []
for i in range(n):
x, y, g, k = map(int, input().split())
a.append([x, y, x + g, y + k])
x, y = map(int, input().split())
ans = -1
# 从后向前遍历地毯
i = n - 1
while i >= 0:
if x >= a[i][0] and x <= a[i][2] and y >= a[i][1] and y <= a[i][3]:
ans = i + 1
break
i -= 1
print(ans)
回文日期
题目链接
a = {1:31, 2:28, 3:31, 4:30, 5:31, 6:30, 7:31, 8:31, 9:30, 10:31, 11:30, 12:31}
# 将年份翻转组合成回文日期
def reverse(n):
date = n
for i in range(4):
date = date * 10 + n % 10
n //= 10
return date
# 检查年月日是否合法
def check(date):
year = date // 10000
if year % 400 == 0 or year % 4 == 0 and year % 100 != 0:
a[2] = 29
else:
a[2] = 28
month = date // 100 % 100;
if month == 0 or month > 12:
return False
day = date % 100
if day == 0 or day > a[month]:
return False
return True
start = int(input())
end = int(input())
ans = 0
# 枚举所有年份
for i in range(0, 10000):
date = reverse(i)
if date >= start and date <= end:
if check(date):
ans += 1
print(ans)
排序
区间k大数查询
题目链接
n = int(input())
a = list(map(int, input().split()))
m = int(input())
for i in range(m):
L, R, k = map(int, input().split())
b = a[L - 1:R]
b.sort(reverse = True)
print(b[k - 1])
桶排序—明明的随机数
题目链接
n = int(input())
a = list(map(int, input().split()))
cnt = [0] * 1010
ans = 0
for x in a:
if cnt[x] == 0:
ans += 1
cnt[x] += 1
print(ans)
for i in range(1, 1001):
if cnt[i] != 0:
print(i, end = " ")
前缀和
前缀和模板
题目链接
n, m = map(int, input().split())
a = list(map(int, input().split()))
s = {0:0}
for i in range(1, n + 1):
s[i] = s[i - 1] + a[i - 1]
for i in range(m):
L, R = map(int, input().split())
print(s[R] - s[L - 1])
字符串
题目链接
字符串编辑
s = input()
a = list(map(str, input().split()))
if a[0] == 'D':
# 将第一个出现的字符替换为空字符
# 注意，replace方法不会改变原字符串的内容
s = s.replace(a[1], "", 1)
elif a[0] == 'I':
index = s.rfind(a[1])
# 将字符串转为列表，进行插入操作
b = list(s)
b.insert(index, a[2])
# 将列表连接为字符串
s = "".join(b)
else:
s = s.replace(a[1], a[2])
print(s)
二分查找
数的范围
题目链接
def lower_bound(L, R, k):
while L < R:
mid = (L + R) // 2
if a[mid] >= k:
R = mid
else:
L = mid + 1
if a[L] == k:
return L
else:
return -1
def upper_bound(L, R, k):
while L < R:
mid = (L + R + 1) // 2
if a[mid] <= k:
L = mid
else:
R = mid - 1
if a[L] == k:
return L
else:
return -1
n, m = map(int, input().split())
a = list(map(int, input().split()))
while m != 0:
m -= 1
k = int(input())
L = lower_bound(0, n - 1, k)
R = upper_bound(0, n - 1, k)
print(L, R)
数学知识
完数
题目链接
n = int(input())
sum = 0
a = 1
# 这里需要每次求出一对约束，优化后，时间复杂度为O(sqrt(n))
while a * a <= n:
if n % a == 0:
sum += a
b = n // a
if b != n and b != a:
sum += b
a += 1
# 注意：特殊处理n为1的情况
if n == 1 or sum != n:
print("no")
else:
print("yes")
```python
## 分解质因数
[题目链接](http://oj.yogeek.cn/problem/lq_cup1058)
```python
# 质数判断
def is_prime(n):
i = 2
while i * i <= n:
if n % i == 0:
return 0
i += 1
else:
return 1
a, b = map(int, input().split())
for n in range(a, b + 1):
# 如果n为质数，直接输出，否则会TLE
if is_prime(n):
print("%d=%d" % (n, n))
continue
#质因数分解
p = 2
print("%d=" % n, end = "")
while n > 1:
while n % p == 0:
print(p, end = "")
n //= p
if n > 1:
print("*", end = "")
else:
print("")
p = p + 1
筛质数
题目链接
# 获取不超过n的所有质数
def get_primes(n):
# 遍历2到n的所有数
for i in range(2, n + 1):
# 如果该数没有被筛除
if st[i] == 0:
# 添加到质数列表中
p.append(i)
# 将i的倍数筛除
j = 2 * i
while j <= n:
st[j] = 1
j = j + i
# 返回质数个数
return len(p)
p = []
n = int(input())
# 标记数组，没有筛除时状态为0
st = [0] * (n + 10)
print(get_primes(n))
回溯算法
阮小二买彩票
题目链接
s = input()
# 将s转为int列表
s = [int(x) for x in s]
n = len(s)
a = []
ans = []
st = [0] * 10
def dfs(t):
# 找到一组序列后，将其转为字符串加入ans列表
if t == n:
ans.append("".join([str(x) for x in a]))
return
for i in range(0, n):
if st[i] == 0:
a.append(s[i])
st[i] = 1
dfs(t + 1)
a.pop()
st[i] = 0
# 回溯生成所有序列
dfs(0)
# 将得到的序列进行排序
ans.sort()
for i in range(len(ans)):
if i == 0:
print(ans[i])
# 如果跟上一个序列不同则输出
elif ans[i] != ans[i - 1]:
print(ans[i])
八皇后问题
题目链接
a = [0] * 20
c = [0] * 20
u = [0] * 20
v = [0] * 20
ans = []
n = 8
def dfs(t):
global n
if t == n:
b = [str(a[i]) for i in range(n)]
ans.append("".join(b))
return
for i in range(n):
if c[i] == 1 or u[i + t] == 1 or v[i - t + n] == 1:
continue
c[i] = u[i + t] = v[i - t + n] = 1
a[t] = i + 1
dfs(t + 1)
c[i] = u[i + t] = v[i - t + n] = 0
dfs(0)
m = int(input())
while m > 0:
n = int(input())
print(ans[n - 1])
m -= 1
黑白皇后问题
题目链接
# 输入
g = []
n = int(input())
ans = 0
for i in range(n):
g.append(list(map(int, input().split())))
# 初始化黑白皇后的列、对角线状态
cw = [0] * (n + 10)
uw = [0] * (n * 3)
vw = [0] * (n * 3)
cb = [0] * (n + 10)
ub = [0] * (n * 3)
vb = [0] * (n * 3)
# 搜索白皇后的情况
def dfs_w(t):
global n
if t == n:
# 成功放置n个白皇后，再搜索黑皇后的情况
dfs_b(0)
return
for i in range(n):
if cw[i] == 1 or uw[i + t] == 1 or vw[i - t + n] == 1:
continue;
if g[t][i] == 0:
continue;
g[t][i] = 0
cw[i] = uw[i + t] = vw[i - t + n] = 1
dfs_w(t + 1)
g[t][i] = 1
cw[i] = uw[i + t] = vw[i - t + n] = 0
# 搜索黑皇后的情况
def dfs_b(t):
global ans
global n
if t == n:
# 成功放置后，方案总数加1
ans += 1
return
for i in range(n):
if cb[i] == 1 or ub[i + t] == 1 or vb[i - t + n] == 1:
continue;
if g[t][i] == 0:
continue;
cb[i] = ub[i + t] = vb[i - t + n] = 1
dfs_b(t + 1)
cb[i] = ub[i + t] = vb[i - t + n] = 0
# 搜索白皇后的情况
dfs_w(0)
print(ans)
贪心
接水问题
题目链接
n, m = map(int, input().split())
a = list(map(int, input().split()))
b = [0] * m
for x in a:
t = min(b)
i = b.index(t)
b[i] += x
print(max(b))
多个队列的接水问题
题目链接
# 计算队列q的等待时间
def wait(q):
sum = 0
n = len(q)
for i in range(n):
sum += q[i] * (n - i - 1)
return sum
n, m = map(int, input().split())
a = list(map(int, input().split()))
# 按打水时间从小到大排序
a.sort()
# 初始化m个队列
q = []
for i in range(m):
q.append([])
# 将n个人安排在m个队列中
for i in range(n):
q[i % m].append(a[i])
# 遍历每个队列，计算等待时间的和
ans = 0
for x in q:
ans += wait(x)
print(ans)
和最大子序列
题目链接
n = int(input())
a = list(map(int, input().split()))
ans = -1e9
sum = 0
for x in a:
if sum < 0:
sum = x
else:
sum += x
ans = max(ans, sum)
print(ans)
递推
N阶楼梯上楼问题
题目链接
n = int(input())
f = [0] * (n + 10)
f[1] = 1
f[2] = 2
for i in range(3, n + 1):
f[i] = f[i - 1] + f[i - 2]
print(f[n])
杨辉三角形
题目链接
n = int(input())
f = []
# 初始化n行杨辉三角
for i in range(n):
f.append([0] * (i + 1))
print(f)
# 计算杨辉三角
i = 0
while i < n:
j = 0
while j <= i:
if(j == 0 or j == i):
f[i][j] = 1
else:
f[i][j] = f[i - 1][j] + f[i - 1][j - 1]
j += 1
i += 1
# 输出杨辉三角
i = 0
while i < n:
j = 0
while j <= i:
print(f[i][j], end = " ")
j += 1
print("")
i += 1
错排问题
题目链接
递推算法题解：错排问题
n = int(input())
f = {}
f[1] = 0
f[2] = 1
for i in range(3, n + 1):
f[i] = (f[i - 1] + f[i - 2]) * (i - 1) % 100003
print(f[n])
动态规划
坐标型动态规划——夺宝奇兵
题目链接
n = int(input())
f = []
for i in range(n):
f.append(list(map(int, input().split())))
# 从山底向上枚举计算
i = n - 2
while i >= 0:
j = 0
while j <= i:
f[i][j] += max(f[i + 1][j], f[i + 1][j + 1])
j += 1
i -= 1
print(f[0][0])