[Hiho] binary and merge sort in reverse order

Describe

We knew Nettle was playing "Battleship これ" on the previous, last, and upper back. After a bitter struggle, nettle again got a lot of boats.
This day nettle is checking its fleet list:

As we can see, the ship's default sort is the rank parameter. But in fact a ship's fire value and the level of the relationship is not big, so there will be a ship than B ship grade, but a ship fire is lower than the case of B ship. For example, the 77-stage dragon to two firepower is less than the 55-level of the XI-Li two.
Now that the nettle will give all the ship's firepower in order of rank, please calculate how many pairs of ships meet the above mentioned situation.

Tip: High firepower is fierce!

Input

Line 1th: an integer n. N indicates the number of ships, 1≤n≤100,000
Line 2nd: n integers, the number of I indicates the fire value of the ship with low rank a[i],1≤a[i]≤2^31-1.

Output

Line 1th: An integer that indicates how many pairs of ships meet "a ship is higher than B ship, but a ship's firepower is lower than ship B."

Sample input

101559614248 709366232 500801802 128741032 1669935692 1993231896 369000208 381379206 962247088 237855491

Sample output

27

Problem Solving Ideas: The merge  sort is to divide the series A[l,h] into two halves a[l,mid] and a[mid+1,h], respectively, and then combine the two halves together. In the process of merging (set L<=i<=mid,mid+1<=j<=h), when the a[i]<=a[j], does not produce reverse order number, when A[i]>a[j], in the first half than the a[i] larger than the number of a[j], will a[j] Put in front of a[i], reverse order number to add mid+1-i. Therefore, you can calculate the number of reverse orders during the merge process in a merged sort.
Note the range, the maximum number of reverse order is n (n-1)/2, the result is saved with a long long.

#include <stdio.h>using namespacestd;Const intN =100000+5;intA[n],tmp[n];Long Longans;voidMerge (intLintMintR) {    inti =l; intj = m +1; intK =l;  while(I <= m && J <=r) {if(A[i] >A[j]) {tmp[k+ +] = a[j++]; Ans+ = M-i +1; }        Else{tmp[k+ +] = a[i++]; }    }     while(i <= m) tmp[k++] = a[i++];  while(J <= r) tmp[k++] = a[j++];  for(inti=l;i<=r;i++) A[i]=tmp[i];}voidMerge_sort (intLintR) {    if(L <r) {intm = (L + r) >>1;        Merge_sort (L,M); Merge_sort (M+1, R);    Merge (L,M,R); }}intMain () {intN; //freopen ("In.txt", "R", stdin);     while(SCANF ("%d", &n)! =EOF) {         for(intI=0; i<n;i++) scanf ("%d",&A[i]); Ans=0; Merge_sort (0, N-1); printf ("%lld\n", ans); }    return 0;}

[Hiho] binary and merge sort in reverse order