[六省联考2017]期末考试

嘟嘟嘟


我看我是老了啊，这zz题都在想暴力（还只搞了20分）……
什么三分， 什么单调性，直接暴力枚举就行了。


我们暴力枚举最后的成绩公布时间，然后算上二分查找(O(logn))时间单次计算就行。
刚开始我一直没想出来，怎么求(n)个数和(x)的差的和，一直在搞什么数据结构，其实直接整体考虑，用(n * x)减去(sum[n])就完事了……
然后计算的时候根据(A, B)的大小关系分两种情况。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
const ull INF = 1e18;
const db eps = 1e-8;
const int maxn = 1e5 + 5;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

ull A, B, C, sumb[maxn], sumt[maxn], ans = INF;
int n, m, t[maxn], b[maxn];

int main()
{
  A = read(), B = read(), C = read();
  n = read(), m = read();
  for(int i = 1; i <= n; ++i) t[i] = read();
  for(int i = 1; i <= m; ++i) b[i] = read();
  sort(t + 1, t + n + 1), sort(b + 1, b + m + 1);
  for(int i = 1; i <= n; ++i) sumt[i] = sumt[i - 1] + t[i];
  for(int i = 1; i <= m; ++i) sumb[i] = sumb[i - 1] + b[i];
  for(int i = 0; i <= b[m]; ++i)
    {
      int pos; ull sum, tp;
      if(B <= A)
	{
	  pos = lower_bound(b + 1, b + m + 1, i) - b;
	  sum = B * (sumb[m] - sumb[pos - 1] - (ull)i * (m - pos + 1));
	}
      else
	{
	  pos = lower_bound(b + 1, b + m + 1, i) - b;
	  ull tot1 = (ull)i * (pos - 1) - sumb[pos - 1];
	  ull tot2 = sumb[m] - sumb[pos - 1] - (ull)i * (m - pos + 1);
	  if(tot1 >= tot2) sum = A * tot2;
	  else sum = A * tot1 + B * (tot2 - tot1);
	}
      pos = lower_bound(t + 1, t + n + 1, i) - t;
      tp = pos ? C * ((ull)i * (pos - 1) - sumt[pos - 1]) : 0;
      ans = min(ans, sum + tp);
    }
  write(ans), enter;
  return 0;
}