BestCoder Round #4 之 Miaomiao's Geometry(2014/8/10)
最后收到邮件说注意小数的问题!此代码并没有过所有数据,请读者参考算法,
自己再去修改一下吧!注意小数问题!
Miaomiao's Geometry
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 10 Accepted Submission(s): 3
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 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.
3
1 2 3
3
1 2 4
4
1 9 100 10
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.
题目大意:
#include <stdio.h>
#include <string.h>
#include <algorithm> using namespace std; int main()
{
int t;
int n;
int i, j;
int a[60];
int b[60], e;
scanf("%d", &t);
while(t--)
{
scanf("%d", &n);
for(i=0; i<n; i++)
{
scanf("%d", &a[i] );
}
sort(a, a+n); e=0;
for(i=1; i<n; i++)
{
b[e++] = a[i] - a[i-1] ;
}
sort(b, b+n-1 ); int max=-1;
int flag; for(i=0; i<n-1; i++)
{
flag=1; //初始化每个间距标记
for(j=1; j<n-1; j++)
{
if(a[j]-b[i]<a[j-1] && a[j]+b[i]>a[j+1] )
{
flag=0;
break;
}
}
if(flag==1)
{
if(b[i] >max )
{
max = b[i] ;
}
}
}
printf("%d", max );
printf(".000\n");
}
return 0;
}
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