BC16&&hdoj5086&&hdoj5087 Revenge of Segment Tree Revenge of LIS II

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概述

Revenge of Segment Tree

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1474 Accepted Submission(s): 526

Problem Description

In computer science, a segment tree is a tree data structure for storing intervals, or segments. It allows querying which of the stored segments contain a given point. It is, in principle, a static structure; that is, its content cannot be modified once the structure is built. A similar data structure is the interval tree.
A segment tree for a set I of n intervals uses O(n log n) storage and can be built in O(n log n) time. Segment trees support searching for all the intervals that contain a query point in O(log n + k), k being the number of retrieved intervals or segments.
---Wikipedia

Today, Segment Tree takes revenge on you. As Segment Tree can answer the sum query of a interval sequence easily, your task is calculating the sum of the sum of all continuous sub-sequences of a given number sequence.

Input

The first line contains a single integer T, indicating the number of test cases.

Each test case begins with an integer N, indicating the length of the sequence. Then N integer Ai follows, indicating the sequence.

[Technical Specification]
1. 1 <= T <= 10
2. 1 <= N <= 447 000
3. 0 <= Ai <= 1 000 000 000

Output

For each test case, output the answer mod 1 000 000 007.

Sample Input

Sample Output


2
20


Hint
For the second test case, all continuous sub-sequences are [1], [2], [3], [1, 2], [2, 3] and [1, 2, 3]. So the sum of the sum of the sub-sequences is 1 + 2 + 3 + 3 + 5 + 6 = 20.
Huge input, faster I/O method is recommended. And as N is rather big, too straightforward algorithm (for example, O(N^2)) will lead Time Limit Exceeded.
And one more little helpful hint, be careful about the overflow of int.

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
#define MOD 1000000007
using namespace std;
const int maxn=10010;
int main()
{
int t;
long long i,n;
scanf("%d",&t);
while(t--){
scanf("%lld",&n);
long long sum=0,num;
for(i=1;i<=n;++i){
scanf("%lld",&num);
sum=(sum+num*(n-i+1)%MOD*i)%MOD;
}
printf("%lldn",sum);
}
return 0;
}

Revenge of LIS II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1332 Accepted Submission(s): 450

Problem Description

In computer science, the longest increasing subsequence problem is to find a subsequence of a given sequence in which the subsequence's elements are in sorted order, lowest to highest, and in which the subsequence is as long as possible. This subsequence is not necessarily contiguous, or unique.
---Wikipedia

Today, LIS takes revenge on you, again. You mission is not calculating the length of longest increasing subsequence, but the length of the second longest increasing subsequence.
Two subsequence is different if and only they have different length, or have at least one different element index in the same place. And second longest increasing subsequence of sequence S indicates the second largest one while sorting all the increasing subsequences of S by its length.

Input

Output

For each test case, output the length of the second longest increasing subsequence.

Sample Input

Sample Output


1
3
2


Hint
For the first sequence, there are two increasing subsequence: [1], [1]. So the length of the second longest increasing subsequence is also 1, same with the length of LIS.

Source

BestCoder Round #16

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
using namespace std;
const int maxn=1010;
int num[maxn];
int dp[maxn][2];
void update(int x,int y,int &temp,int &ans){
if(x>=temp){
ans=temp;temp=x;
}
else if(x>ans)ans=x;
if(y>=temp){
ans=temp;temp=y;
}
else if(y>ans)ans=y;
}
int main()
{
int t,i,j,k,n;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
for(i=1;i<=n;++i){
scanf("%d",&num[i]);
}
int temp=0,ans=0;
for(i=1;i<=n;++i){
dp[i][0]=1;dp[i][1]=0;
for(j=1;j<i;++j){
if(num[i]>num[j]){
int cnt1=dp[j][0]+1;
int cnt2=dp[j][1]+1;
if(cnt1>dp[i][0])swap(dp[i][0],cnt1);
if(cnt1>dp[i][1])dp[i][1]=cnt1;
if(cnt2>dp[i][0])swap(dp[i][0],cnt2);
if(cnt2>dp[i][1])dp[i][1]=cnt2;
}
}
update(dp[i][0],dp[i][1],temp,ans);
}
printf("%dn",ans);
}
return 0;
}