Please give me idea ?

start from the right, when we encounter element x, the answer is query from [0..x-1], then update the segment tree by increasing value on index x by 1 on the segment tree.

this is similar to counting sort, if the maximum element is too large, just compress the data first.

You can use Binary Indexed Tree. Here is a code.

void update(int i) { for(; i < maxNumber; i += i&-i) BIT[i]++; }

int find(int i) { int res(0); for(; i; i -= i&-i) res += BIT[i]; return res; }

void solve() { read(A); for(i = N; i >= 1; i--) { r[i] = find(A[i]); update(A[i]); } } You can update BIT( Binary Indexed Tree ) int O(logN) and get the result in O(logN). Then complexity is O(NlogN) but it is faster than segment tree and you can implement this easier.

We can solve this problem by using BIT and Segment Tree (of course). The condination is the values must be smaller than 10000000 and strictly larger than 0 (if not, you can compress your array by replace value of array with rank of it in array, for example, [500,700,500,850] -> [1,2,1,3]). In our segment tree, node k whick manage segment [l,r] will contains sum of elements from l to r. Let i run from n down to 1, with i, we assign result[i] := segmenttree.get(1,a[i]-1), then call segmenttree.set(a[i], 1) (increase element a[i]-th in tree by 1). Lastly, we print array result[].

This is pseudo code which I hope it can help you:

declare tr : SegmentTree;
:::
for i = n downto 1
    rslt[i]=tr.get(1, a[i]-1);
    tr.set(a[i], 1);

print result[]

It is easier when you use Binary Indexed Tree.