Dylans loves sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

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

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<math.h>
using namespace std;
const int N = 1005;
#define LL __int64

int n,q,a[N],l,r,dp[N][N];
int main()
{
    while(scanf("%d%d",&n,&q)>0){
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i]);
        memset(dp,0,sizeof(dp));
        for(int k=1; k<n; k++)
        for(int i=1; i+k<=n; i++){
            int j=i+k;
            dp[i][j]=dp[i][j-1]+dp[i+1][j]-dp[i+1][j-1];
            if(a[i]>a[j])
                dp[i][j]++;
        }
        while(q--){
            scanf("%d%d",&l,&r);
            printf("%d\n",dp[l][r]);
        }
    }
    return 0;
}