Sort(归并)

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
代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int MAXN = 1010;
int num[MAXN], temp[MAXN]; 
int ans;
void mg(int l, int m, int r){
    int i = l, j = m + 1, cur = l;
    while(i <= m && j <= r){
        if(num[i] < num[j]){
            temp[cur++] = num[i++];
        }
        else{
            ans += j - cur;
            temp[cur++] = num[j++];
        }
    }
    for(int i = l; i <= r; i++)
        num[i] = temp[i];
}
void ms(int l, int r){
    int m;
    if(l < r){
        m = (l + r) >> 1;
        ms(l, m);
        ms(m + 1, r);
        mg(l, m, r);
    }
}
int main(){
    int T,n;
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        for(int i = 0; i < n; i++){
            scanf("%d", num + i);
        }
        ans = 0;
        ms(0, n - 1);
        printf("%d
", ans);
    }
    return 0;
}