NUC_TeamTEST_C && POJ2299（只有归并）

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstring>
 4 #define LL long long　　　　　　　　　　　　　　//LL 代替 long long 的写法 中间数据会超出　　int

 5 using namespace std;
 6 
 7 const int max_size = 500010;
 8 
 9 int arry[max_size], tmp_arry[max_size];
10 
11 LL Merge(int *arr, LL beg, LL mid, LL end, int *tmp_arr)
12 {
13     memcpy(tmp_arr+beg, arr+beg, sizeof(int)*(end-beg+1));
14     LL i = beg;
15     LL j = mid + 1;
16     LL k = beg;
17     LL inversion = 0;
18     while(i <= mid && j <= end)
19     {
20         if(tmp_arr[i] <= tmp_arr[j])   ///如果合并逆序数的时候，前边小于等于后边，就不用记录逆序数的值
21         {
22             arr[k++] = tmp_arr[i++];
23         }else{
24             arr[k++] = tmp_arr[j++];   ///如果不是，则要记录逆序数的值
25             inversion += (mid - i + 1);///简单画下就能看出
26         }
27     }
28 
29     while(i <= mid)                    ///把没有并入arr数组的数并入
30 
31     {
32         arr[k++] = tmp_arr[i++];
33     }
34     while(j <= end)
35     {
36         arr[k++] = tmp_arr[j++];
37     }
38     return inversion;
39 }
40 
41 LL MergeInversion(int *arr, LL beg, LL end, int *tmp_arr)
42 {
43     LL inversions = 0;
44     if(beg < end)
45     {
46         LL mid = (beg + end) >> 1;
47         inversions += MergeInversion(arr, beg, mid, tmp_arr);    ///分成两段分别进行记录，递归的进行下去，找逆序数和
48         inversions += MergeInversion(arr, mid+1, end, tmp_arr);
49         inversions += Merge(arr, beg, mid, end, tmp_arr);
50     }
51     return inversions;
52 }
53 
54 int main()
55 {
56     LL n;
57 
58     while(cin >> n)
59     {
60         if(n == 0)
61             break;
62         for(int i = 0; i < n; ++i)
63             scanf("%d", &arry[i]);
64         memcpy(tmp_arry, arry, sizeof(int)*n);
65         cout << MergeInversion(arry, 0, n-1, tmp_arry) << endl;
66     }
67     return 0;
68 }