D. Once Again...
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given an array of positive integers a1, a2, ..., an × T of length n × T. We know that for any i > n it is true that ai = ai - n. Find the length of the longest non-decreasing sequence of the given array.
Input
The first line contains two space-separated integers: n, T (1 ≤ n ≤ 100, 1 ≤ T ≤ 107). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 300).
Output
Print a single number — the length of a sought sequence.
Examples
input
Copy
4 3 3 1 4 2
output
5
Note
The array given in the sample looks like that: 3, 1, 4, 2, 3, 1, 4, 2, 3, 1, 4, 2. The elements in bold form the largest non-decreasing subsequence.
求n*t这么长的最长不递减子序列
数据太大 直接跑肯定GG 但是 序列如果是n*n已经包含了所有的可能，所以只需要求出来n*n的最长+(t-n)*一个n中出现最多的即可。
可以想象n*n最长中加塞了很多一样的--->(t-n)*最多的
long long a[10005];
int main ()
{
long long n,t;
scanf("%lld%lld",&n,&t);
for(int i=1;i<=n;i++)
scanf("%lld",&a[i]);
long long ans=0;
for(int i=1;i<=n;i++)
{
long long sum=1;
for(int j=i+1;j<=n;j++)
{
if(a[i]==a[j])
sum++;
}
ans=max(ans,sum);
}
long long len[10005];
memset(len,0,sizeof(len));
int l;
if(t>=n)
l=n*n;
else
l=t*n;
for(int i=n+1;i<=l;i++)
{
a[i]=a[i-n];
}
for(int i=1;i<=l;i++)
{
len[i]=1;
for(int j=1;j<i;j++)
{
if(a[i]>=a[j])
len[i]=max(len[i],len[j]+1);
}
}
long long ans1=0;
if(t<n)
{
for(int i=1;i<=l;i++)
{
// printf("%d %lld\n",i,ans1);
ans1=max(ans1,len[i]);
}
}
else
{
ans1=(t-n)*ans;
long long temp=0;
for(int i=1;i<=n*n;i++)
temp=max(temp,len[i]);
ans1=ans1+temp;
}
printf("%lld\n",ans1);
return 0;
}