Problem:
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The Number of Inversions

For a given sequence $A = \{a_0, a_1, ... a_{n-1}\}$, the number of pairs $(i, j)$ where $a_i > a_j$ and $i < j$, is called the number of inversions. The number of inversions is equal to the number of swaps of Bubble Sort defined in the following program:

bubbleSort(A)
cnt = 0 // the number of inversions
for i = 0 to A.length-1
for j = A.length-1 downto i+1
if A[j] < A[j-1]
swap(A[j], A[j-1])
cnt++

return cnt

For the given sequence $A$, print the number of inversions of $A$. Note that you should not use the above program, which brings Time Limit Exceeded.

Input

In the first line, an integer $n$, the number of elements in $A$, is given. In the second line, the elements $a_i$ ($i = 0, 1, .. n-1$) are given separated by space characters.

output

Print the number of inversions in a line.

Constraints

$ 1 \leq n \leq 200,000$

$ 0 \leq a_i \leq 10^9$

$a_i$ are all different

Sample Input 1

5
3 5 2 1 4

Sample Output 1

6

Sample Input 2

3
3 1 2

Sample Output 2

2

Solution in C (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int cmp(const void *const a, const void *const b) {
    return *(int *) a - *(int *) b;
}

int read(int *t, int n) {
    int r = 0;
    while (n) {
        r += t[n];
        n -= (n & -n);
    }
    return r;
}

int update(int *t, int n, int s, int v) {
    while (n < s) {
        t[n] += v;
        n += (n & -n);
    }
}

int main() {
    int n, *a, *b, *c;
    scanf("%d", &n);
    a = malloc(sizeof(int) * n);
    b = malloc(sizeof(int) * n);
    int sz = 1;
    while (sz < n) sz *= 2;
    sz += 1;
    c = malloc(sizeof(int) * sz);
    memset(c, 0, sizeof(int) * sz);
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    memcpy(b, a, sizeof(int) * n);
    qsort(b, n, sizeof(int), cmp);
    for (int i = 0; i < n; i++) a[i] = (int) ((int *) bsearch(&a[i], b, n, sizeof(int), cmp) - b);
    long long r = n;
    for (int i = 0; i < n; i++) {
        update(c, a[i] + 1, sz, 1);
        r += i - read(c, a[i] + 1);
    }
    printf("%lld\n", r);
}