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BestCoder Round #4 Miaomiao's Geometry (暴力)
Problem Description
There are N point on X-axis . Miaomiao would like to cover them ALL by using segments with same length.
There are 2 limits:
1.A point is convered if there is a segments T , the point is the left end or the right end of T.
2.The length of the intersection of any two segments equals zero.
For example , point 2 is convered by [2 , 4] and not convered by [1 , 3]. [1 , 2] and [2 , 3] are legal segments , [1 , 2] and [3 , 4] are legal segments , but [1 , 3] and [2 , 4] are not (the length of intersection doesn‘t equals zero), [1 , 3] and [3 , 4] are not(not the same length).
Miaomiao wants to maximum the length of segements , please tell her the maximum length of segments.
For your information , the point can‘t coincidently at the same position.
There are 2 limits:
1.A point is convered if there is a segments T , the point is the left end or the right end of T.
2.The length of the intersection of any two segments equals zero.
For example , point 2 is convered by [2 , 4] and not convered by [1 , 3]. [1 , 2] and [2 , 3] are legal segments , [1 , 2] and [3 , 4] are legal segments , but [1 , 3] and [2 , 4] are not (the length of intersection doesn‘t equals zero), [1 , 3] and [3 , 4] are not(not the same length).
Miaomiao wants to maximum the length of segements , please tell her the maximum length of segments.
For your information , the point can‘t coincidently at the same position.
Input
There are several test cases.
There is a number T ( T <= 50 ) on the first line which shows the number of test cases.
For each test cases , there is a number N ( 3 <= N <= 50 ) on the first line.
On the second line , there are N integers Ai (-1e9 <= Ai <= 1e9) shows the position of each point.
There is a number T ( T <= 50 ) on the first line which shows the number of test cases.
For each test cases , there is a number N ( 3 <= N <= 50 ) on the first line.
On the second line , there are N integers Ai (-1e9 <= Ai <= 1e9) shows the position of each point.
Output
For each test cases , output a real number shows the answser. Please output three digit after the decimal point.
Sample Input
3 3 1 2 3 3 1 2 4 4 1 9 100 10
Sample Output
1.000 2.000 8.000HintFor the first sample , a legal answer is [1,2] [2,3] so the length is 1. For the second sample , a legal answer is [-1,1] [2,4] so the answer is 2. For the thired sample , a legal answer is [-7,1] , [1,9] , [10,18] , [100,108] so the answer is 8.
终于结果仅仅可能出现两种情况,长度为某个区间长度,或为区间长度的一半。枚举每一个长度,仅仅要符合条件就更新最大值。
#include <stdio.h> #include <string.h> #include <algorithm> #include <math.h> #include <stack> #define lson o<<1, l, m #define rson o<<1|1, m+1, r using namespace std; typedef long long LL; const int maxn = 1500; const int MAX = 0x3f3f3f3f; const int mod = 1000000007; int t, n; double a[55]; int ok(double cur) { int vis = 0; for(int i = 2; i < n ; i++) { double l, r; if(vis == 0) l = a[i]-a[i-1]; else l = a[i]-a[i-1]-cur; if(l >= cur) vis = 0; else { r = a[i+1]-a[i]; if(r > cur ) vis = 1; else if(r == cur) { vis = 0; i++; } else return 0; } } return 1; } int main() { scanf("%d", &t); while(t--) { scanf("%d", &n); for(int i = 1; i <= n; i++) scanf("%lf", &a[i]); sort(a+1, a+1+n); double tmp ,ans = 0; for(int i = 2; i <= n; i++) { tmp = a[i]-a[i-1]; if(ok(tmp)) ans = max(ans, tmp); tmp = (a[i]-a[i-1])/2; if(ok(tmp)) ans = max(ans, tmp); } printf("%.3lf\n", ans); } return 0; }
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BestCoder Round #4 Miaomiao's Geometry (暴力)
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