# Codeforces 657c-bear and contribution [idea title]

Source: Internet
Author: User

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The special thing about the topic is that only \$+1\$ \$+5\$ these two operations we have to consider how to use this condition

When you think about it, you can see that if the target value of the optimal solution is \$x (\$ to increase the value of at least \$k\$ personal to \$x) \$

Then there must be a man whose initial value is within the range of \$[x-4, x]\$

Otherwise, the \$x\$ minus the \$5\$ can get a better solution.

Therefore, the target value that may be the optimal solution is only \$n * 5\$ species

Now consider how to quickly calculate an answer for each target value in the context of \$O (N) \$ enumeration of target values

My first idea was to write the \$5\$ in terms of the remainder of the \$mod 5\$ to record the prefixes and whatever

However such two-point answer has a \$log \$ two fractional set subscript another \$log \$

Although the topic gave \$4s\$ should be able to pass but always feel that this is not elegant

Re-analysis of the condition of the problem we can find that each query is actually the minimum distance from the target \$k \$ number and this \$k\$ is unchanged

So now the problem is to maintain a data structure that supports finding the smallest number of \$k in a container and adding a number

This is obviously a heap of total complexity is \$O (n + nlogk) \$

The sort is not the same as the cost of transferring all the initial values to the target value in the case of the 5\$ \$5\$ \$mod

So we're just going to keep \$5\$ a heap.

`1#include <cstdio>2#include <cstring>3#include <cmath>4#include <algorithm>5#include <queue>6 using namespacestd;7 Const intN = 2e5 +Ten;8 Const Long LongINF =1e9;9 intA[n], Dis[n *5];Ten Long Longsum[5]; One Long LongAns =1e18; A intN, K, B, C, CNT; -Priority_queue <Long Long> q[5]; - intMain () the { -scanf"%d%d%d%d", &n, &k, &b, &c); -b = min (b, c *5); -      for(inti =1; I <= N; ++i) +     { -scanf"%d", &a[i]); +          for(intj =0; J <5; ++j) Adis[cnt++] = A[i] +J; at     } -Sort (A +1, A +1+n); -Sort (dis, dis +CNT); -CNT = unique (dis, dis + cnt)-dis; -     intnow =1; -      for(inti =0; I < CNT; ++i) in     { -         intx = Dis[i], kind = (inf + x)%5; to         Long LongCost ; +          while(now <= n && A[now] <=x) -         { the              for(intj =0; J <5; ++j) *             { \$Cost = (INF + j-a[now])/5* B + (INF + j-a[now])%5*C;Panax Notoginseng Q[j].push (cost); -SUM[J] + =Cost ; the                 if((int) Q[j].size () >k) +                 { ASUM[J]-=q[j].top (); the Q[j].pop (); +                 } -             } \$++Now ; \$         } -         Long Longy = (inf + kind-x)/5* b *K; -         if((int) q[kind].size () = =k) theans = min (ans, sum[kind)-y); -     }Wuyiprintf"%lld\n", ans); the     return 0; -}`

Codeforces 657c-bear and contribution [idea title]

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