一、第一题：鱼塘钓鱼

解题思路：多路归并+优先队列
        首先枚举能走到的距离然后再用优先队列将最大的值累加

【Python程序代码】

from heapq import *
n = int(input())
a = [0] + list(map(int,input().split()))
b = [0] + list(map(int,input().split()))
l = [0,0] + list(map(int,input().split()))
for i in range(2,n+1):l[i] +=l[i-1]
t = int(input())
ans = 0
i = 1
def gt(s,j):return s//j+1
while i<=n and t-l[i]>=0:new_t = t - l[i]res,hq = 0,[]for j in range(1,i+1):ci = gt(a[j],b[j])for k in range(ci):heappush(hq,-max(a[j]-k*b[j],0))for k in range(new_t):if hq:res -= heappop(hq)else:breakans = max(ans,res)i += 1
print(ans)

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二、第二题：技能升级

 

解题思路：多路归并+二分
        对攻击力大小进行二分，ck函数是检查能否升级这么多次。

【Python程序代码】

n,m = map(int,input().split())
a,b = [],[]
for i in range(n):a_,b_ = map(int,input().split())a.append(a_); b.append(b_)
l,r = 0,10000000def ck(mid):res = 0for i in range(n):if a[i]<mid:continueres += (a[i]-mid)//b[i] + 1if res>=m:return Truereturn False
while l<r:mid = (l+r+1)>>1if ck(mid):l = midelse:r = mid-1
cnt,res=m,0
for i in range(n):if a[i]<r:continuet = (a[i]-r)//b[i] +1cnt -= tres += (2*a[i]-(t-1)*b[i])*t//2
print(res+cnt*r)

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三、第三题：谦虚的数字

解题思路： 多路归并
        一个谦虚数一定是由比它小的谦虚数乘一个数得到。

【Python程序代码】

from math import *
k,n = map(int,input().split())
a = list(map(int,input().split()))
st = [0]*(k+10)
res = [1]
for _ in range(n):ans = inffor i in range(k):p = res[ st[i] ]*a[i]if p<ans:ans=pfor i in range(k):p = res[ st[i] ]*a[i]if p==ans:st[i]+=1res.append(ans)
print(res[-1])

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四、第四题：序列

 解题思路：多路归并
       两组两组之间合并，对其中一个进行排序，构造出可以多路归并的数组。

【Python程序代码】

from heapq import *
T = int(input())
def work(a,b,n):hq,c = [],[0]*(n+5)for i in range(n):heappush(hq,(a[0]+b[i],0))for i in range(n):x,y = heappop(hq)c[i] = xheappush(hq,(x-a[y]+a[y+1],y+1))for i in range(n):a[i]=c[i]for _ in range(T):m,n = map(int,input().split())a = list(map(int,input().split()))a.sort(); a = a + [0]*100for i in range(m-1):b = list(map(int,input().split()))work(a,b,n)for i in range(n):print(a[i],end=' ')print()