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The Number of Inversions

For a given sequence A={a0,a1,...an−1}A={a0,a1,...an−1}, the number of pairs (i,j)(i,j) where ai>ajai>aj and i<ji<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 AA, print the number of inversions of AA. Note that you should not use the above program, which brings Time Limit Exceeded.

Input

In the first line, an integer nn, the number of elements in AA, is given. In the second line, the elements aiai (i=0,1,..n−1i=0,1,..n−1) are given separated by space characters.

output

Print the number of inversions in a line.

Constraints

• 1≤n≤200,0001≤n≤200,000
• 0≤ai≤1090≤ai≤109
• aiai are all different

Sample Input 1

53 5 2 1 4

Sample Output 1

6

Sample Input 2

33 1 2

Sample Output 2

2

这道题的坑点就是，下标以及统计次数sum都要用long类型，而不能用int，不然会WA，我感觉很奇怪，有的题20w的数据int可以放的下，有的却不行，希望大家知道原因的告诉我。

#include<iostream>using namespace std;typedef long long ll;const long maxn=200005;const ll maxv=1e9+5;long sum=0;void Merge(ll*a,long left,long mid,long right){    long n1=mid-left,n2=right-mid;    ll *L,*R;    L=new ll [n1+1];R=new ll [n2+1];    for(long i=0;i<n1;i++)        L[i]=a[left+i];    for(long j=0;j<n2;j++)        R[j]=a[mid+j];    L[n1]=R[n2]=maxv;    long m=0,n=0;    for(long i=left;i<right;i++)    {        if(L[m]<=R[n]){a[i]=L[m];m++;}        else        {            sum+=n1-m;            a[i]=R[n];            n++;        }    }}void Merge_Sort(ll*a,long left,long right){    if(left<right-1)    {        long mid=(left+right)/2;        Merge_Sort(a,left,mid);        Merge_Sort(a,mid,right);        Merge(a,left,mid,right);    }}int main(){    long n;    ll a[maxn];    cin>>n;    for(long i=0;i<n;i++)        cin>>a[i];    Merge_Sort(a,0,n);    cout<<sum<<endl;    return 0;}