A.归并排序

归并排序 （求逆序数）

归并排序：递归+合并+排序

时间复杂度：O（n logn）    空间复杂度：O（n）

用途：1.排序  2.求逆序对数

Description

In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence9 1 0 5 4 ,        Ultra-QuickSort produces the output0 1 4 5 9 .        Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.       

Input

The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.      

Output

For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.      

Sample Input

5
9
1
0
5
4
3
1
2
3
0

Sample Output

6
0

题意：
给出长度为n的无序序列，每次只能交换相邻的两个数，求至少要交换多少次才使得该序列为递增序列。

分析：
1.根据时间范围估计时间复杂度，归并排序的时间复杂度一般为O（n logn）
2.算法步骤： 1.划分问题：把序列分为元素个数尽量相等的两半
           2.递归求解：把两半元素分别排序
           3.合并问题：把两个有序表合成一个
3.保存时，用long long 类型，否则会wa的

代码：

 #include<cstdio>
 #include<iostream>
 using namespace std;
 const int maxn=;

 int a[maxn],b[maxn];
 long long ans;

 void merge(int l,int m,int r)       //归并排序
 {
     int i=l,j=m+,t=;
     while(i<=m&&j<=r)   //从小到大排序
     {
         if(a[i]>a[j])
         {
             b[t++]=a[j++];
             ans+=m-i+;
         }
         else b[t++]=a[i++];
     }
     while(i<=m)
         b[t++]=a[i++];
     while(j<=r)
         b[t++]=a[j++];
     for(int i=;i<t;i++)   //合并
         a[l+i]=b[i];
 }

 void merge_sort(int l,int r)  //划分问题
 {
     if(l<r)
     {
         int m=(l+r)/;
         merge_sort(l,m);
         merge_sort(m+,r);
         merge(l,m,r);
     }
 }

 int main()
 {
     int n;
     while(scanf("%d",&n)!=EOF)
     {
         if(n==)
             break;
         ans=;
         for(int i=;i<n;i++)
             scanf("%d",&a[i]);
         merge_sort(,n-);
         printf("%lld\n",ans);
     }
     return ;
 }

本来想着完全不看别人的代码都由自己做，可是看了一天了只是有了思路，还是写不出来。。。又看来了别人的代码。。。