HDU 5273 Dylans loves sequence

Problem Description

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

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

直接树状数组预处理出所有情况，然后输出即可

#include<iostream>
#include<cstdio>
#include<vector>
#include<iostream>
#include<queue>
#include<cstdlib>
#include<map>
using namespace std;
const int maxn = 1005;
const int low(int x){ return x&-x; }
int f[maxn];
int a[maxn], b[maxn], c[maxn][maxn];
map<int, int> M;
int n, q, tot, l, r;

void add(int x)
{
    for (int i = x; i <= tot; i += low(i)) f[i]++;
}

int sum(int x)
{
    int ans = 0;
    for (int i = x; i; i -= low(i)) ans += f[i];
    return ans;
}

int main()
{
    while (scanf("%d%d", &n, &q) != EOF)
    {
        for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i - 1] = a[i];
        sort(b, b + n);
        for (int i = tot = 0; i < n; i++) M[b[i]] = ++tot;
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= tot; j++) f[j] = 0;
            c[i][i] = 0;    add(M[a[i]]);
            for (int j = i + 1; j <= n; j++)
            {
                c[i][j] = c[i][j - 1] + j - i - sum(M[a[j]]);
                add(M[a[j]]);
            }
        }
        while (q--)
        {
            scanf("%d%d", &l, &r);
            printf("%d\n", c[l][r]);
        }
    }
    return 0;
}