Codeforces Round #FF (Div. 2) C. DZY Loves Sequences
2014-07-17 13:55
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题目: DZY Loves Sequences
给你n 个数, 让你至多可以改变一个数的情况下,求得最长的严格递增子序列的长度。
智商捉急。 。 首先预处理出来对于每个位置i, 以这个位置为左端点的最长序列的长度, 和以他为右端点的最长序列的长度。
之后遍历一遍,判断每个位置是否改变,求得max。
code:
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
#define N 100005
#define INF 1000000000
int n, num
, l
, r
, ans;
int main()
{
//freopen("in.txt", "r", stdin);
scanf("%d", &n);
for(int i=1; i<=n; i++)
{
scanf("%d", &num[i]);
}
num[0]= -100;
num[n+1]= INF;
for(int i=1; i<=n; i++)
{
if(num[i] > num[i-1]) l[i] = l[i-1]+1;
else l[i] = 1;
}
for(int i=n; i>=1; i--)
{
if(num[i]<num[i+1]) r[i] = r[i+1] + 1;
else r[i] = 1;
}
ans=0;
for(int i=1; i<=n; i++)
{
ans = max(ans, max(r[i], l[i]));
if(i>1 && num[i] <= num[i-1]) ans = max(ans, l[i-1] +1 );
if(i<n && num[i] >= num[i+1]) ans = max(ans, r[i+1] +1);
if(i>1 && i<n && num[i-1]+1 < num[i+1]) ans = max(ans, l[i-1] + r[i+1] +1);
}
printf("%d\n", ans);
return 0;
}
给你n 个数, 让你至多可以改变一个数的情况下,求得最长的严格递增子序列的长度。
智商捉急。 。 首先预处理出来对于每个位置i, 以这个位置为左端点的最长序列的长度, 和以他为右端点的最长序列的长度。
之后遍历一遍,判断每个位置是否改变,求得max。
code:
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
#define N 100005
#define INF 1000000000
int n, num
, l
, r
, ans;
int main()
{
//freopen("in.txt", "r", stdin);
scanf("%d", &n);
for(int i=1; i<=n; i++)
{
scanf("%d", &num[i]);
}
num[0]= -100;
num[n+1]= INF;
for(int i=1; i<=n; i++)
{
if(num[i] > num[i-1]) l[i] = l[i-1]+1;
else l[i] = 1;
}
for(int i=n; i>=1; i--)
{
if(num[i]<num[i+1]) r[i] = r[i+1] + 1;
else r[i] = 1;
}
ans=0;
for(int i=1; i<=n; i++)
{
ans = max(ans, max(r[i], l[i]));
if(i>1 && num[i] <= num[i-1]) ans = max(ans, l[i-1] +1 );
if(i<n && num[i] >= num[i+1]) ans = max(ans, r[i+1] +1);
if(i>1 && i<n && num[i-1]+1 < num[i+1]) ans = max(ans, l[i-1] + r[i+1] +1);
}
printf("%d\n", ans);
return 0;
}
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