HDU 5273 Dylans loves sequence reverse order number simple recursion

Dylans loves sequence

Time Limit:20 Sec

Memory limit:256 MB

Topic Connection

http://acm.hdu.edu.cn/showproblem.php?pid=5273

Descriptiondylans gets the N number A[1]...a[n].
There are q questions, each problem shaped like (l,r)
He needs to ask for the number of reverse l−r in these numbers.
More formally, he needs to ask for a two-tuple (x, y) number, making l≤x,y≤r and X<y and A[x]>a[y]

Input

The first line has two numbers n and Q.
The second line gives N numbers a[1]...a[n].
The next Q line, each line gives two numbers L, R.

n≤1000,q≤100000,l≤r,1≤a[i]≤231−1

Output

For each query, the output is in reverse order.

Sample Input

3 2
3 2 1
1 2
1 3

Sample Output

1
3

HINT

Test instructions

Exercises

N is only 1000, so I want to how to come.
The easiest thing to think about is the start of the enumeration, and then the Nlog (N) time to calculate the inverse pairs, maintained with a tree array.
(unfortunately BC does not give card ...) Woo Woo)
Think carefully about how easy it is to do N2.
Set Ans[l][r] to the number of l∼r in reverse order. First of all, we have a violent first count of ANS[1][1..N].
Then I from the 2∼n enumeration, each time computes the reverse order starting from I to.
So what Ans[i][j] less than ans[i−1][j]? Yes, less a[i−1] The contribution of this number.
We'll open another accumulator, CNT. Enumeration J from I∼n, if a[i−1] and a[j] produce reverse pairs on the cnt[j]=−1
Then we add CNT from right to left (because contributions are prefixes and properties)
Last Ans[i][j]=ans[i−1][j]+cnt[j].
After preprocessing all the answers you can ask O (1).

Code

#include <cstdio>#include<cmath>#include<cstring>#include<ctime>#include<iostream>#include<algorithm>#include<Set>#include<vector>#include<sstream>#include<queue>#include<typeinfo>#include<fstream>#include<map>#include<stack>typedefLong Longll;using namespacestd;//freopen ("d.in", "R", stdin);//freopen ("D.out", "w", stdout);#defineSspeed ios_base::sync_with_stdio (0); Cin.tie (0)#defineTest Freopen ("Test.txt", "R", stdin)#defineMAXN 2000001#defineMoD 10007#defineEPS 1e-9Const intinf=0x3f3f3f3f;Constll infll =0x3f3f3f3f3f3f3f3fll;inline ll Read () {ll x=0, f=1;CharCh=GetChar ();  while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();}  while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; ch=GetChar ();} returnx*F;}//**************************************************************************************intd[1005][1005];intsum[1005][1005];inta[1005];intMain () {intN=read (), q=read ();  for(intI=1; i<=n;i++) A[i]=read ();  for(intI=1; i<=n;i++)    {         for(intj=i+1; j<=n;j++)        {            if(a[j]<A[i]) d[i][j]=d[i][j-1]+1; Elsed[i][j]=d[i][j-1]; }         for(intj=i-1; j>=1; j--)        {            if(a[j]>A[i]) d[i][j]=d[i][j+1]+1; ElseD[i][j]=d[i][j+1]; }    }     for(intI=1; i<=n;i++)    {         for(intj=i;j<=n;j++) Sum[i][j]=sum[i][j-1]+D[j][i]; }     while(q--)    {        intX=read (), y=read (); printf ("%d\n", Sum[x][y]); }}

HDU 5273 Dylans loves sequence reverse order number simple recursion