Dylans loves sequence(hdu5273)

Dylans loves sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 385    Accepted Submission(s): 193

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 <string.h>
int ans[1010][1010],s[1010],cnt[1010];
int main()
{
    int n,m,i,j,sum,l,r;
    scanf("%d %d",&n,&m);
    for(i=1;i<=n;i++)
    {
        scanf("%d",&s[i]);
    }
    for(i=1;i<=n;i++)
    {
        sum=0;
        for(j=1;j<=i;j++)
        {
            if(s[i]<s[j])
            {
                sum++;
            }
        }
        ans[1][i]=ans[1][i-1]+sum;
    }
    for(i=2;i<=n;i++)
    {
        memset(cnt,0,sizeof(cnt));
        for(j=i;j<=n;j++)
        {
            if(s[i-1]>s[j])
            {
                cnt[j]=cnt[j-1]-1;
            }
            else
            {
                cnt[j]=cnt[j-1];
            }
            ans[i][j]=ans[i-1][j]+cnt[j];
        }
    }
    while(m--)
    {
        scanf("%d %d",&l,&r);
        printf("%d
",ans[l][r]);
    }
    return 0;
}