EOJ2018.10 month (B math + thinking problem)

Portal: Problem B

Https://www.cnblogs.com/violet-acmer/p/9739115.html

Test instructions

　　Find the smallest sequence s with sub-sequence A, and the sequence S is the p -Moganshan sequence.

Exercises

It is easy to think of P = max_ai+1, and a[1] corresponds to s[1], otherwise it needs to be in a[1] before adding other numbers to make a[1]-> a s[i], certainly longer than the non-additive sequence.

Traversal a[] Array, divided into three cases of discussion

①a[i] > A[i-1]

In this case, a[i-1]--a[i] is a continuous sequence in s, res + = A[i]-a[i-1].

②a[i] < a[i-1]

This condition corresponds to the case of A[i], P-and a[i-1], res + = P-a[i]+a[i-1].

③a[i] = = A[i-1]

Equal, just a period of difference, res + = P.

Note that the result will be a long long type.

AC Code:

1#include <iostream>2#include <cstdio>3 using namespacestd;4 #definell Long Long5 Const intmaxn=2e5+ -;6 7 intA[MAXN];8 9 intMain ()Ten { One     intN; Ascanf"%d",&n); -      for(intI=1; I <= n;++i) -scanf"%d", A +i); thell res=1; -     intp=0; -      for(intI=1; I <= n;++i) -p=Max (p,a[i]); +p++; -      for(intI=2; I <= n;++i) +     { A         if(A[i] > a[i-1]) atRes + = a[i]-a[i-1]; -         Else if(A[i] < a[i-1]) -         { -             intk=a[i-1]/p; -Res + = ((k +1) *p-a[i-1])+A[i]; -         } in         Else -Res + =p; to     } +printf"%lld\n", res); -}

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EOJ2018.10 month (B math + thinking problem)