【322 sort 归并排序】

Sort

时间限制：1000 ms  |  内存限制：65535 KB

难度：4

描述

You want to processe a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. Then how many times it need.
For example, 1 2 3 5 4, we only need one operation : swap 5 and 4.

输入

The input consists of T number of test cases.(<0T<1000) Each case consists of two lines: the first line contains a positive integer n (n <= 1000); the next line contains a permutation of the n integers from 1 to n.

输出

For each case, output the minimum times need to sort it in ascending order on a single line.

样例输入

2
3
1 2 3
4
4 3 2 1

样例输出

0
6

/*归并排序，记录交换的次数
在a[i]>a[j]  需要交换的次数就为  mid-i+1
*/

#include<cstdio>
int a[100005],temp[100005];
int memgesort(int a[],int first,int mid,int last){
    int i=first,j=mid+1,n=mid,m=last,s=first,count=0;
    while(i<=n&& j<=m){
        if(a[i]<=a[j])
            temp[s++]=a[i++];
        else{
            temp[s++]=a[j++];
            count+=(mid-i+1);
        }
    }
    while(i<=n)
        temp[s++]=a[i++];
    while(j<=m)
        temp[s++]=a[j++];
    for(int i=first;i<=last;i++)
        a[i]=temp[i];
    return count;
}
int memsort(int a[],int first,int last){
    int mid=(first+last)/2;
    int x,y,z;
    if(first>=last)
        return 0;
    x=memsort(a,first,mid);
    y=memsort(a,mid+1,last);
    z=memgesort(a,first,mid,last);
    return (x+y+z);
}
int main(){
    int m,n;
    scanf("%d",&m);
    while(m--){
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
    printf("%d\n",memsort(a,0,n-1));
    }
}