二分搜索 2015百度之星初赛1 HDOJ 5248 序列变换

题目传送门

 /*
     二分搜索：在0～1e6的范围找到最小的max (ai - bi)，也就是使得p + 1 <= a[i] + c or a[i] - c
     比赛时以为是贪心，榨干智商也想不出来:(
 */
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <cmath>
 #include <iostream>
 using namespace std;

 const int MAXN = 1e5 + ;
 const int INF = 0x3f3f3f3f;
 int a[MAXN];
 int n;

 bool check(int c)
 {
     int p = -1e6;
     for (int i=; i<=n; ++i)
     {
         int now = max (p + , a[i] - c);
         if (now > a[i] + c)    return false;
         p = now;
     }

     return true;
 }

 int main(void)        //2015百度之星初赛1 1003 序列变换
 {
     int t, cas = ;
     scanf ("%d", &t);
     while (t--)
     {
         scanf ("%d", &n);
         for (int i=; i<=n; ++i)    scanf ("%d", &a[i]);

         int l = , r = 1e6;
         while (l < r)
         {
             int mid = (l + r) >> ;
             if (check (mid))    r = mid;
             else    l = mid + ;
         }

         printf ("Case #%d:\n", ++cas);
         printf ("%d\n", l);
     }

     return ;
 }

 /*
 2
 2
 1 10
 3
 2 5 4
 */