HDU 5273 (Recursive)

Dylans loves sequenceaccepts:250submissions:806Time limit:2000/1000 MS (java/others)Memory limit:131072/131072 K (java/others) Problem description

Dylans got the
    
       N    
  
 
    
 Number
   
    
        a[1]. . . a [N] 
 

    
   。 Yes
    
        Q   
  
 
    
 Problems, each of which is like a
    
        (L,R)    
  
 
    
 He needs to ask the
    
        L? R  
  
 
    
 The number of inverse pairs in these numbers. More formally, he needs to ask for a two-tuple
    
        (x,y)   
  
 
    
 The number that makes
    
        L≤x,y≤R   
  
 
    
 And
    
        x<y     
  
 
    
 And
   
    
        a[x]>a[y  ]
 

    
   

Enter a description

The first line has two numbers.
    
       N    
  
 
    
 And
    
        Q   
  
 
    
 。 The second line gives
    
       N    
  
 
    
 A number
   
    
        a[1]. . . a [N] 
 

    
   。 The next
    
        Q   
  
 
    
 Rows, given two numbers per line
    
        L,R    
  
 
    
 。
   
    N  ≤ +,Q≤100000,L≤R,1≤a[I]≤  2  ?1
   

Output description

For each query, the output is in reverse order.

Input sample

3 23 2 11 21 3

Output sample

1 3

Hint

There should be no extra space at the end of each line read in the hack data.

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RES[I][J] (j<=i) represents the number of I in the 1 to J in this interval to the inverse of the contribution. GAO[I][J] for the inverse logarithm of the interval [i,j], how to quickly find out the inverse logarithm of the interval, considering the contribution of the number of K (I=<K<=J), the contribution of the k number should be 1 to K contribution minus 1 to i-1 contribution, then by gao[i][j-1] can O ( 1) to find out the gao[i][j]. After preprocessing, the query is O (1).

#include <bits/stdc++.h>using namespace Std;typedef long long ll;const int maxn = 1e3+; #define foreach (It,v) for ( __typeof (V.begin ()) it = V.begin (); It! = V.end (); ++it) #define REP (i,a,b) for (int i= (a);i< (b); i++) #define PB push_backint A[maxn];ll RES[MAXN][MAXN],GAO[MAXN][MAXN]    ; int main (int argc, char const *argv[]) {int n,q;        while (scanf ("%d%d", &n,&q) ==2) {for (int i = 1; I <= N; i++) {scanf ("%d", a+i);            } for (int i = 1; I <= N; i++) {res[i][0] = 0;        for (int j = 1; J <= I; j + +) Res[i][j] = Res[i][j-1] + (A[j] > A[i]);            } for (int i = 1; I <= N; i++) {gao[i][i] = 0;            for (int j = i + 1; j <= N; j + +) {Gao[i][j] =gao[i][j-1] + res[j][j]-res[j][i-1];            }} while (q--) {int l,r;scanf ("%d%d", &l,&r);        printf ("%i64d\n", Gao[l][r]); }} return 0;}

HDU 5273 (Recursive)