二分搜索 2015百度之星初赛1 HDOJ 5248 序列变换

 

题目传送门

 1 /*  2  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  3  比赛时以为是贪心，榨干智商也想不出来:(  4 */  5 #include <cstdio>  6 #include <algorithm>  7 #include <cstring>  8 #include <cmath>  9 #include <iostream> 10 using namespace std; 11 12 const int MAXN = 1e5 + 10; 13 const int INF = 0x3f3f3f3f; 14 int a[MAXN]; 15 int n; 16 17 bool check(int c) 18 { 19 int p = -1e6; 20 for (int i=1; i<=n; ++i) 21  { 22 int now = max (p + 1, a[i] - c); 23 if (now > a[i] + c) return false; 24 p = now; 25  } 26 27 return true; 28 } 29 30 int main(void) //2015百度之星初赛1 1003 序列变换 31 { 32 int t, cas = 0; 33 scanf ("%d", &t); 34 while (t--) 35  { 36 scanf ("%d", &n); 37 for (int i=1; i<=n; ++i) scanf ("%d", &a[i]); 38 39 int l = 0, r = 1e6; 40 while (l < r) 41  { 42 int mid = (l + r) >> 1; 43 if (check (mid)) r = mid; 44 else l = mid + 1; 45  } 46 47 printf ("Case #%d:\n", ++cas); 48 printf ("%d\n", l); 49  } 50 51 return 0; 52 } 53 54 /* 55 2 56 2 57 1 10 58 3 59 2 5 4 60 */