Mathematics:Ultra-QuickSort(POJ 2299)

　　　　　　　　　　　　　　　　　　

　　　　　　　　　　　　　　　　　　极度快速排序

　　题目大意：在一个输入数组中找逆序数。。。

　　水题，求逆序数的很好的算法，就是MergeSort，和我之前发的DNA那个差不多，最后就是后台数据很大，答案要用long long

　　

 #include <iostream>
 #include <functional>
 #include <algorithm>

 using namespace std;

 static int num[], tmp[];
 typedef int Position;

 void Merge_Sort(Position, Position, long long *const);
 void Merge(Position,Position, Position, long long *const);

 int main(void)
 {
     int length;
     long long ans;
     while (~scanf("%d", &length))
     {
         if (length == )break;
         for (int i = ; i < length; i++)
             scanf("%d", &num[i]);
         ans = ;
         Merge_Sort(, length - , &ans);
         printf("%lld\n", ans);
     }
     return ;
 }

 void Merge_Sort(Position Left, Position Right, long long *const reverse_num)
 {
     if (Left < Right)
     {
         Position Mid = (Left + Right) / ;
         Merge_Sort(Left, Mid, reverse_num);
         Merge_Sort(Mid + , Right, reverse_num);
         Merge(Left, Mid, Right, reverse_num);
     }
 }

 void Merge(Position Left, Position mid, Position Right, long long *const reverse_num)
 {
     Position lpos = Left, rpos = mid + , lend = mid, rend = Right, pos = Left;

     while (lpos <= lend && rpos <= rend)
     {
         if (num[lpos] <= num[rpos])
             tmp[pos++] = num[lpos++];
         else
         {
             (*reverse_num) += lend - lpos + ;
             tmp[pos++] = num[rpos++];
         }
     }
     while (lpos <= lend)
         tmp[pos++] = num[lpos++];
     while (rpos<=rend)
         tmp[pos++] = num[rpos++];
     for (pos = Left; pos <= Right; pos++)
         num[pos] = tmp[pos];
 }