| ||||||||||
| Online Judge | Problem Set | Authors | Online Contests | User | ||||||
|---|---|---|---|---|---|---|---|---|---|---|
| Web Board Home Page F.A.Qs Statistical Charts | Current Contest Past Contests Scheduled Contests Award Contest | |||||||||
给点测试数据把,不知道错哪了我的思路,线性扫秒,保存当前的最大和最小,以及他们的下标,并计算出当前距离,如果遇到一个比max大的,那么将起赋为最小同时改变下标,动态的更新最长距离
#include<iostream>
using namespace std;
int a[50001];
int main()
{
int n,i,max,min,sum,len,ti,tj;
while(scanf("%d",&n)!=EOF)
{
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
max=-999999999;
min=999999999;
ti=1;tj=1;
len=0;
sum=0;
for(i=1;i<=n;i++)
{
// cout<<"case:"<<i<<endl;
if(a[i]>max)
{
max=a[i];
tj=i;
}
if(a[i]<min)
{
min=a[i];
ti=i;
}
len=tj-ti;
if(len>sum)sum=len;
if(a[i]<max)
{
min=a[i];
ti=i;
max=a[i];tj=i;
}
// cout<<"i="<<ti<<" j="<<tj<<endl;
// cout<<"min="<<min<<" max="<<max<<endl;
// system("pause");
}
if(sum==0)printf("-1\n");
else printf("%d\n",sum);
}
return 0;
}
Followed by:
Post your reply here: |
All Rights Reserved 2003-2013 Ying Fuchen,Xu Pengcheng,Xie Di
Any problem, Please Contact Administrator