标签：des style color os io for ar cti
3 3 1 2 3 3 1 2 4 4 1 9 100 10
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.
最终结果只可能出现两种情况，长度为某个区间长度，或为区间长度的一半，枚举每个长度，只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;
}
BestCoder Round #4 Miaomiao's Geometry （暴力）,布布扣,bubuko.com
BestCoder Round #4 Miaomiao's Geometry （暴力）
标签：des style color os io for ar cti
原文地址：http://blog.csdn.net/u013923947/article/details/38481555
