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BestCoder Round #4 Miaomiao's Geometry （暴力）

2024-09-15 20:28:40   219人阅读

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.
 


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.
 


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
8.000

Hint
For 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&#39;s Geometry （暴力）


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