牛客寒假训练营第 5 场题解

B 题智乃的瓷砖

考点：暴力

思路：判断 i, j 的奇偶性即可。

import sys
import math
input = sys.stdin.readline
n, m = map(int, input().split())
a = [['' for _ in range(m)] for _ in range(n)]
for i in range(n):
    for j in range(m):
        if i % 2 == 0:
            if j % 2 == 0:
                a[i][j] = '/'
            else:
                a[i][j] = '\\'
        else:
            if j % 2 == 0:
                a[i][j] = '\\'
            else:
                a[i][j] = '/'
for i in range(n):
    for j in range(m):
        print(a[i][j], end='')
    print()

D 题智乃的果子

考点：优先队列，贪心

思路：读题可想到肯定是从小往大合并会比较优。我们先用优先队列记录一下果子的重量和数量，每次都取最轻的处理，把其中的一半两两合并；如果这一组只有一个果子，我们就再取一组与它合并。

 sys
 collections  deque
 = sys.stdin.readline
mod = ** + 
n = (())
mp = {}
total = 
 _  (n):
    c, w = (, ().split())
    mp[w] = mp.get(w, ) + c
    total += c

a = deque((mp.items()))
b = deque()

 ():
     a  a[][] == :
        a.popleft()
     b  b[][] == :
        b.popleft()

 ():
    clean()
      b  (a  a[][] <= b[][]):
        w, c = a[]
         c == :
            a.popleft()
        :
            a[] = (w, c - )
         w
    :
        w, c = b[]
         c == :
            b.popleft()
        :
            b[] = (w, c - )
         w

 ():
    clean()
      b  (a  a[][] < b[][]):
        w = a[][]
    :
        w = b[][]
    c = 
     a  a[][] == w:
        c += a[][]
        a.popleft()
     b  b[][] == w:
        c += b[][]
        b.popleft()
     w, c

 ():
     c <= :
        
     b  b[-][] == w:
        b[-] = (w, b[-][] + c)
    :
        b.append((w, c))

ans = 
 total > :
    w, c = pop_all()
    k = c // 
     k:
        ans = (ans + (k % mod) * (( * w) % mod)) % mod
        total -= k
        push( * w, k)
     c & :
        x = pop()
        s = w + x
        ans = (ans + s) % mod
        total -= 
        push(s, )
(ans % mod)