Codeforces Round #200 (Div. 2) E. Read Time(二分)

题目链接

这题，关键不是二分，而是如果在t的时间内，将n个头，刷完这m个磁盘。

看了一下题解，完全不知怎么弄。用一个指针从pre，枚举m，讨论一下。只需考虑，每一个磁盘是从右边的头，刷过来的（左边来的之前刷了）。

思维是硬伤。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <vector>
 #include <cmath>
 #include <algorithm>
 using namespace std;
 #define LL __int64
 LL p[],h[];
 int n,m;
 int judge(LL x)
 {
     int pre = ;
     LL temp;
     int i;
     for(i = ;i < n;i ++)
     {
         if(abs(p[pre]-h[i]) > x) continue;
         if(p[pre] < h[i])
         temp = h[i] + max((x-(h[i]-p[pre]))/,x-*(h[i]-p[pre]));
         else
         temp = h[i] + x;
         while(p[pre] <= temp&&pre < m)pre ++;
     }
     if(pre == m)
     return ;
     else
     return ;
 }
 LL bin()
 {
     LL str,end,mid;
     str = ;
     end = 100000000000000ll;
     while(str < end)
     {
         mid = (str + end)/;
         if(judge(mid))
         end = mid;
         else
         str = mid + ;
     }
     return str;
 }
 int main()
 {
     int i;
     scanf("%d%d",&n,&m);
     for(i = ;i < n;i ++)
     {
         scanf("%I64d",&h[i]);
     }
     for(i = ;i < m;i ++)
     {
         scanf("%I64d",&p[i]);
     }
     printf("%I64d\n",bin());
     return ;
 }