Find the number of reverse order

To find the time limit of reverse order: -Ms | Memory Limit:65535KB Difficulty:5

Describe

In one arrangement, if the front and back positions of a pair of numbers are opposite to the size order, that is, the previous number is greater than the subsequent number, then they are called an inverse. The total number of reverse order in a permutation is called the inverse number of the permutation.

Now, to give you a sequence of n elements, you have to figure out how many reverse-order it is.

For example, 1 3 2 reverse number is 1.

Input

the first line enters an integer T that represents the number of groups of test data (1<=t<=5)
Each row of test data for each group is an integer n representing a total of n elements in the sequence (2〈=n〈=1000000)
The following line has a total of n integer Ai (0<=ai<1000000000), representing all the elements in the sequence.

Data assurance in multiple sets of test data, more than 100,000 numbers of test data have a maximum of one set.

Output

output the inverse number of the sequence

Sample input

221 131 3 2

Sample output

01

Code 1,

 #include <stdio.h>long long A[1000005];long long t[1000005]; Long Long count;void Merg (Long long *a,int left,int Right,long long *t) {if (right-left>1) {int mid=left+ (right-left)/2; int P=left,q=mid,i=left;merg (a,left,mid,t); Merg (a,mid,right,t); while (p<mid| | Q<right) {if (q>=right| |  (P<mid&&a[p]<=a[q]))  T[i++]=a[p++];else{t[i++]=a[q++];count=count+mid-p;} }for (i=left;i<right;i++) {a[i]=t[i];}}} int main (void) {int t,n;scanf ("%d", &t), while (t--) {scanf ("%d", &n), for (int i=0;i<n;i++) {scanf ("%lld", &a[i]);} Count=0;merg (a,0,n,t);p rintf ("%lld\n", count);} return 0;} 
 
 
 

Code 2,

 #include <stdio.h>long long A[1000005];long long t[1000005]; Long Long count;void Merg (Long long *a,int left,int Right,long long *t) {if (right-left>1) {int mid=left+ (right-left)/2; int L=left,r=mid,i=left;merg (a,left,mid,t), Merg (A,mid,right,t), while (L<mid&&r<right) {if (a[l]<=  A[R]) t[i++]=a[l++];else{t[i++]=a[r++];count=count+mid-l;} }while (L<mid) {t[i++]=a[l++];} while (r<right) {t[i++]=a[r++];count=count+mid-l;} for (i=left;i<right;i++) {a[i]=t[i];}}} int main (void) {int t,n;scanf ("%d", &t), while (t--) {scanf ("%d", &n), for (int i=0;i<n;i++) {scanf ("%lld", &a[i]);} Count=0;merg (a,0,n,t);p rintf ("%lld\n", count);}        return 0;} 
 
 
 

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Find the number of reverse order