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 <assert.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>

static long long int merge(int A[], int left, int mid, int right) {
    int n1 = mid - left;
    int n2 = right - mid;
    int i, j, k;
    int *L, *R;
    long long int c = 0;

    L = malloc(sizeof(int) * n1 + sizeof(int));
    R = malloc(sizeof(int) * n2 + sizeof(int));
    for ( i = 0; i < n1; i++ ) {
        L[i] = A[left + i];
    }
    L[i] = INT_MAX;
    for ( i = 0; i < n2; i++ ) {
        R[i] = A[mid + i];
    }
    R[i] = INT_MAX;    
    
    for ( i = 0, j = 0, k = left; k < right; k++ ) {
        if ( L[i] <= R[j] ) {
            A[k] = L[i];
            i++;
        } else {
            A[k] = R[j];
            c += n1 - i;        /* 飛び越えたR側の個数 */
            j++;
        }
    }
    free(L);
    free(R);

    return c;
}

static long long int MergeSort(int A[], int left, int right) {
    long long int c = 0;
    
    if ( left + 1 >= right ) {
        return 0;
    }

    int mid = (left + right) / 2;
    c += MergeSort(A, left, mid);
    c += MergeSort(A, mid, right);
    c += merge(A, left, mid, right);

    return c;
}

int main(void) {
    int * A;
    int n;
    long long int c;

    scanf("%d\n", &n);
    A = malloc(sizeof(int) * n);

    for ( int i = 0; i < n; i++ ) {
        int a;
        scanf("%d", &a);
        A[i] = a;
        (void)fgetc(stdin);
    }

    c = MergeSort(A, 0, n);

    printf("%lld\n", c);

#if 0
    for (int i = 0; i < n; i++ ) {
        printf("%d ", A[i]);
    }
    printf("\n");
#endif

    return 0;
}