Dylans loves sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 405    Accepted Submission(s): 201

Problem Description

Dylans is given  N  numbers  a[1]....a[N]

And there are  Q  questions.

Each question is like this  (L,R)

his goal is to find the “inversions” from number  L  to number  R .

more formally,his needs to find the numbers of pair（ x,y ）,
that  L≤x,y≤R  and  x<y  and  a[x]>a[y]

 

Input

In the first line there is two numbers  N  and  Q .

Then in the second line there are  N  numbers: a[1]..a[N]

In the next  Q  lines,there are two numbers  L,R  in each line.

N≤1000,Q≤100000,L≤R,1≤a[i]≤231−1

 

Output

For each query,print the numbers of "inversions”

 

Sample Input

   
   
   
   

    
    
    
    3 2
3 2 1
1 2
1 3
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1
3


    
    
    
    
 
     
 
      Hint 
     
 You shouldn't print any space in each end of the line in the hack data. 
    
    
    
    
     
   
   
   
   

 

Source

BestCoder Round #45

 

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#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<stack>
#include<queue>
using namespace std;
int dp[1010][1010];
int figure[1010];
int cont;
int main()
{
    int m,n;
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&figure[i]);
        }
        for(int i=2;i<=n;i++)       //从1到2.3.4.。。。。n的逆序数；
        {
            dp[1][i]=dp[1][i-1];
            for(int j=1;j<i;j++)
            {
                if(figure[j]>figure[i])
                dp[1][i]++;
            }
        }
        for(int i=2;i<n;i++)                                       由1到n推i到j；
        {
            cont=figure[i-1]>figure[i]?1:0;
            for(int j=i+1;j<=n;j++)
            {
                if(figure[i-1]>figure[j])cont++;
                dp[i][j]=dp[i-1][j]-cont;
            }
        }
        int u,v;
        while(m--)
        {
            scanf("%d%d",&u,&v);
            printf("%d\n",dp[u][v]);
        }

    }
    return 0;
}