BestCoder Round #4 Miaomiao & #39; s Geometry (violent ),

BestCoder Round #4 Miaomiao's Geometry (violent ),
Problem DescriptionThere are N point on X-axis. Miaomiao wowould 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.

InputThere 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.

OutputFor each test cases, output a real number shows the answser. Please output three digit after the decimal point.

Sample Input

331 2 331 2 441 9 100 10

 

Sample Output

1.0002.0008.000HintFor 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.

The final result may only show two conditions: the length is the length of a certain interval, or half of the Interval Length. If each length is enumerated, the maximum value is updated as long as the conditions are met.

#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, rusing 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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