蓝桥·算法赛—

文章目录

小白入门赛
- 1-元宵节快乐
- 2-猜灯谜
- 3-数学奇才
- 4-你不干？有的是帕鲁干！
- 5-等腰三角形
- 6-计算方程
强者挑战赛
- 1-你不干？有的是帕鲁干！
- 2-等腰三角形
- 3-计算方程
- 4-谁是帕鲁？
- 5-源石开采
- 6-集合划分

小白入门赛

1-元宵节快乐

2-猜灯谜

n = int(input())
a = list(map(int, input().split()))
for i in range(n):left = i - 1 if i > 0 else -1right = i + 1 if i < n-1 else 0print(a[left] + a[right], end=" ")

3-数学奇才

n = int(input())
a = list(map(int, input().split()))
# 当负数个数小于等于总数一半时,可以将 1 ~ i、1 ~ i - 1 翻转# 总次数为 2 * 负数个数,不超过 n
# 当负数超过一半,先将所有数翻转,再按上述方法,# 总次数为 2 * 小于一半 + 1,不超过 n
print(sum(i if i > 0 else -i for i in a))

4-你不干？有的是帕鲁干！

T = int(input())
for _ in range(T):x = int(input())if x % 8 == 0 and x != 0:print("Yes")print("{0} {1}".format(x // 4 - 1, x // 4 + 1))else:print("No")

5-等腰三角形

N = int(input())
A = sorted(list(map(int, input().split())))
B = sorted(list(map(int, input().split())))pa = 0; pb = 0; ans = 0
while pa < N and pb < N:if 2 * A[pa] > B[pb]:pa += 1; pb += 1ans += 1else:pa += 1
print(ans)

6-计算方程

from math import sqrt, log, floort = int(input())
for _ in range(t):k, m = map(int, input().split())left = 1; right = m ** 2while left <= right:mid = (left + right) // 2num = sqrt(mid) + floor(log(mid, k))if num > m:right = mid - 1else:left = mid + 1print(left)

强者挑战赛

1-你不干？有的是帕鲁干！

T = int(input())
for _ in range(T):x = int(input())if x % 8 == 0 and x != 0:print("Yes")print("{0} {1}".format(x // 4 - 1, x // 4 + 1))else:print("No")

2-等腰三角形

N = int(input())
A = sorted(list(map(int, input().split())))
B = sorted(list(map(int, input().split())))pa = 0; pb = 0; ans = 0
while pa < N and pb < N:if 2 * A[pa] > B[pb]:pa += 1; pb += 1ans += 1else:pa += 1
print(ans)

3-计算方程

from math import sqrt, log, floort = int(input())
for _ in range(t):k, m = map(int, input().split())left = 1; right = m ** 2while left <= right:mid = (left + right) // 2num = sqrt(mid) + floor(log(mid, k))if num > m:right = mid - 1else:left = mid + 1print(left)

4-谁是帕鲁？

5-源石开采

def build(o, L, R): # 建树if L == R: # 划分到一个区间Tree[o][0], Tree[o][1] = a[L], -1returnMid = (L + R) >> 1 # 划分build(o << 1, L, Mid) # o 的左端点和管辖区间build(o << 1 | 1, Mid + 1, R) # o 的右端点和管辖区间Tree[o][0] = max(Tree[o << 1][0], Tree[o << 1 | 1][0])Tree[o][1] = max(Tree[o << 1][1], Tree[o << 1 | 1][1])Tree[o][1] = max(Tree[o][1], min(Tree[o << 1][0], Tree[o << 1 | 1][0]))def query(left, right, o, L, R): # 返回 [left, right] 中的最大值和第二大值if left <= L and R <= right: # 当 [left, right] 完全覆盖 [L, R]return Tree[o]Mid = (L + R) >> 1 # 划分区间if right <= Mid: # 全在左边return query(left, right, o << 1, L, Mid)elif left > Mid: # 全在右边return query(left, right, o << 1 | 1, Mid + 1, R)else: # Mid 在 [left, right] 中间temp1 = query(left, right, o << 1, L, Mid) # 左边temp2 = query(left, right, o << 1 | 1, Mid + 1, R) # 右边ans = [max(temp1[0], temp2[0]), max(max(temp1[1], temp2[1]), min(temp1[0], temp2[0]))]return ans # 返回一个数组n, q = map(int, input().split())
a = [0] + list(map(int, input().split()))
Tree = [[0, 0] for _ in range(n << 2)] # 线段树（第一大, 第二大）
build(1, 1, n) # 1.建树（结点编号, a 左端点, a 右端点）
cnt = 0 # 统计源石数量
for _ in range(q):left, right = map(int, input().split())# （查询区间左端点, 查询区间右端点, 结点编号, 结点管辖左端点, 结点管辖右端点）cnt += sum(query(left, right, 1, 1, n))
m = int(input()) # 博弈论取石头
print(cnt, "\nred" if cnt % (m + 1) else "\nblue")