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  • AGC041

    B

    降序排序

    • (ile p)(i)一定合法
    • (i>p)
      (a_i+m<a_{p})则显然不合法
      那么([1,p)[i,n])这些全选。至于(xin[p,i))中间这些点,最多升至(a_i+m),由于(a_xge a_i),其不会消耗完天数,故边界条件就是将其全部升至(a_i+m)

    C

    (n=5)

    aabc.
    ..bcd
    fee.d
    f.ghh
    iigjj
    

    (n=7)

    aabbcc.
    ddee..c
    ffgg..c
    ....geb
    ....geb
    ....fda
    ....fda
    

    (3|n)
    (3 imes 3)的对角线拼起来

    (2|n)
    大概是这样

    aa....cd
    bb....cd
    cdaa....
    cdbb....
    ..cdaa..
    ..cdbb..
    ....cdaa
    ....cdbb
    

    然后(n)不是(3)倍数的奇数,(7 imes 7)之后是偶数块

    D

    做法1
    合法的充要条件

    • (a_1le a_2le cdotsle a_{n-1}le a_n)
    • (forall kin[1,n),.s.t~sumlimits_{i=1}^{k+1} a_i>sumlimits_{i=n-k+1}^n a_iLongrightarrow sumlimits_{i=1}^{k+1} a_i>sumlimits_{i=n-k+1}^n a_i(k=leftlfloorfrac{n-1}{2} ight floor))

    (2|n),令(k=leftlfloorfrac{n-1}{2} ight floor)

    • (a_{k+2}-a_ile a_{k+2}-a_{i-1}(iin(k+2,1)))(a_{i}-a_{k+2}le a_{i+1}-a_{k+2}(iin(k+2,n)))
    • (sumlimits_{i=1}^{k+1} a_i>sumlimits_{i=n-k+1}^n a_iLongrightarrow a_{k+2}>(sumlimits_{i=n-k+1}^n a_i-a_{k+2})+(sumlimits_{i=1}^{k+1}a_{k+2}-a_i))

    (f_{i,s}(i<k+2))为前(i)个数,与(a_{k+2})差的和为(s)
    为满足差的单调性,考虑分层转移,即在最外层降序枚举与(a_{k+2})的差(l),然后内层(f_{i+1,s+l}=f_{i+1,s+l}+f_{i,s})。我们有(ille n)。故复杂度为(O(n^2logn))
    (g_{i,s})为后(i)个数,与(a_{k+2})的差的和为(s)。转移类似

    (n)不是(2)的倍数,令(k=leftlfloorfrac{n-1}{2} ight floor)

    • (a_{k+1}-a_ile a_{k+1}-a_{i-1}(iin(k+1,1)))(a_{i}-a_{k+1}le a_{i+1}-a_{k+1}(iin(k+1,n)))
    • (sumlimits_{i=1}^{k} a_i>sumlimits_{i=n-k+1}^n a_iLongrightarrow a_{k+1}>(sumlimits_{i=n-k+1}^n a_i-a_{k+1})+(sumlimits_{i=1}^{k}a_{k+1}-a_i))

    做法2
    (a_i=1+sum limits_{j=1}^i b_i)

    • (sum b_i<n,b_ige 0)
    • (forall kin[1,n),.s.t~sumlimits_{i=1}^{k+1} (1+sum limits_{j=1}^i b_i)>sumlimits_{i=n-k+1}^n (1+sum limits_{j=1}^i b_i)(k=leftlfloorfrac{n-1}{2} ight floor))

    第二个限制可以化简成(sumlimits_{i=1}^{n}c_ib_ile 0)。其中(c_i)可以分(n)的奇偶性得到。特殊的,其中只有(c_1)为负数,且为(-1)
    即约数条件可以重新写为

    • (b_1le n-1-sumlimits_{i=2}^n b_i)
    • (b_1ge sumlimits_{i=2}^n c_ib_i)

    在已知(b_i(iin[2,n]))的条件下,合法的(b_1)的个数为(max(0,n-sumlimits_{i=2}^n (c_i+1)b_i))
    (f_j)(sumlimits_{i=2}^n (c_i+1)b_i)=j)的方案数,(ans=sumlimits_{i=0}^{n-1}(n-i)f_i)

    E

    这题比较简单

    F

    这题题解有点难写

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  • 原文地址:https://www.cnblogs.com/Grice/p/13271760.html
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