[ACM] beg number reverse order

Time Limit: 2 Sec Memory Limit: 64 MB
Submit: 544 Solved: 78

Title Description

In one arrangement, if the longitudinal position of a pair of numbers in reverse order of magnitude, i.e. the number greater than the number in front of the latter, they are called a reverse order. A reverse arrangement on the total number of the called number in reverse order.
Now, to give you a sequence of N elements, you determine the number of reverse it is.
For example inverse number of 1132 it is.

Entry

Each row represents a first line of input test data an integer number of T (1 <= T <= 5) test data for each group is an integer N represents the number of columns in a total of N elements (2 <= N <= 1000000) followed by a total of one row of N integers Ai (0 <= Ai <1000000000), represents the number of all elements in the column. A plurality of sets of data to ensure that the test data, the test data of more than 100,000 at most a number of groups.

Export

Reverse order of the number of output number sequences

Sample input

2
2
1 1
3
1 3 2

Sample Output

0
1

#include<stdio.h>
#include<string.h>
int a[1000005],b[1000005];
long long count;
void merge(int low,int mid,int high){
    int i=low,j=mid+1,k=low;
    while((i<=mid)&&(j<=high)){
        if(a[i]<=a[j]){
            b[k++]=a[i++];
        }
        else{
            b[k++]=a[j++];
            count+=mid-i+1;
        }
    }
    while(i<=mid)
        b[k++]=a[i++];
    while(j<=high)
        b[k++]=a[j++];
    for(int i=low; i<=high; i++)
        a[i]=b[i];
}
void sort(long long low,long long high){
    int mid=(low+high)/2;
    int x,y,z;
    if(low>=high)
        return ;
    sort(low,mid);
    sort(mid+1,high);
    merge(low,mid,high);
    return ;
 
}
int main(){
    int N;
    scanf("%d",&N);
    while(N--){
        int n;
        count=0;
        scanf("%d",&n);
        for(int i=0; i<n; i++)
            scanf("%d",&a[i]);
        sort(0,n-1);
        printf("%lld\n",count);
    }
    return 0;
}