BestCoder Round #4 Miaomiao's Geometry （暴力），

BestCoder Round #4 Miaomiao's Geometry （暴力），
Problem DescriptionThere 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. 

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.

最終結果只可能出現兩種情況，長度為某個區間長度，或為區間長度的一半，枚舉每個長度，只要符合條件就更新最大值。

#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, 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;}