Codeforces Round #446 (Div. 2)

A. Greed

题意

问能否把所有可乐倒到两个罐子里

题解

挑两个最大的罐子

 1 sum = 0
 2 b = []
 3 n = int(input())
 4 line1 = input().split()
 5 line2 = input().split()
 6 for i in range(n):
 7     sum += int(line1[i])
 8     b.append(int(line2[i]))
 9     continue
10 b.sort()
11 cap = b[n-1] + b[n-2]
12 if sum > cap:
13     print('NO')
14 else:
15     print('YES')
View Code

B. Wrath

题意

每个人手里都有一根长度为L[i]的EX咖喱棒, 只要铃声一响, 就把自己前面L[i]个人砍死(同时), 问最后活下来多少人

题解

把自己能砍死的最远的人标记+1, 自己标记-1,  扫一遍标记数组, 是正数就是死人, 0就是活人

 1 n = int(input())
 2 shit = [0 for i in range(n + 1)]
 3 line = input().split()
 4 for i in range(n):
 5     claw = int(line[i])
 6     shit[i] -= 1
 7     pre = i - claw
 8     if pre < 0:
 9         pre = 0
10     shit[pre] += 1
11     continue
12 ans = n
13 tmp = 0
14 for i in range(n):
15     tmp += shit[i]
16     if tmp > 0:
17         ans -= 1
18     continue
19 print(ans)
View Code

C. Pride

题意

给定一个数组a, 有一种操作可以把相邻的两个数(x, y)中的一个(x或者y)替换为gcd(x, y), 问全部替换成1的最少操作, 不可能时输出'-1'

题解

数组a内有1的时候, 显然答案是非1的个数

否则, 只要找到最快生成一个1的操作次数

从a[i]开始, 计算a[i]到a[j]的gcd, (i < j < n), 当区间gcd为1时, 更新最小操作数为区间的长度, 即j - i + 1

 1 import math
 2 n = int(input())
 3 a = []
 4 k = 0
 5 line = input().split()
 6 for i in range(n):
 7     a.append(int(line[i]))
 8     if a[i] != 1:
 9         k += 1
10     continue
11 ans = n
12 for i in range(n):
13     temp = a[i]
14     cur = 0
15     for j in range(i + 1, n):
16         cur += 1
17         temp = math.gcd(temp, a[j])
18         if temp == 1:
19             if ans > cur:
20                 ans = cur
21                 pass
22             pass
23         continue
24     continue
25 if k == 0:
26     print('0')
27 elif ans == n:
28     print('-1')
29 else:
30     print(ans + k - 1)
View Code

D. Gluttony

 1 n = int(input())
 2 a = [int(i) for i in input().split()]
 3 if n == 1:
 4     print(a[0])
 5 else:
 6     a_with_index = [[a[i], i] for i in range(n)]
 7     a_with_index.sort()
 8     ans = [0 for i in range(n)]
 9     for i in range(n):
10         ans[a_with_index[i][1]] = str(a_with_index[(i + 1) % n][0])
11     print(" ".join(ans))
View Code

E. 

posted @ 2017-11-18 09:34  Backlight8891  阅读(282)  评论(0)  编辑  收藏  举报