HDU 1394 minimum inversion number [number of reverse orders]

Question link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1394

Give N, and then give a group of 0 ~ N-1, calculate the number of reverse orders of this sequence, and the minimum value of the number of reverse orders after the exchange. The exchange rules are as follows:

A1, A2,..., An-1, an (where m = 0-the initial seqence)
A2, A3,..., An, A1 (where M = 1)
A3, A4,..., An, A1, A2 (where m = 2)
...
An, A1, A2,..., An-1 (where M = N-1)

For example, if we have obtained 4 3 2 1 0, the number of reverse orders is 10, then

4 switch to the end 3 2 1 0 4 for element 4, the number of reverse orders is reduced by 4 first, then 4-4, and the last number of reverse orders is 6

3 switch to the end 2 1 0 4 3 for element 3, the reverse order number first reduces 3, then increases 4-3, and the last Reverse Order Number is 4

Then, the minimum value of the reverse order is obtained through a loop.

// Sort the data from the merge statement to the reverse sequence # include <stdio. h> int g_ncount; void mergearray (int A [], int first, int mid, int last, int temp []) {int I = first, j = Mid + 1; int M = mid, n = last; int K = 0; while (I <= M & J <= N) // number after a [I] Number A [J] {if (a [I] <A [J]) temp [k ++] = A [I ++]; else {temp [k ++] = A [J ++]; // A [J] and each of the preceding numbers can form a reverse order number pair g_ncount + = m-I + 1 ;}} while (I <= m) temp [k ++] = A [I ++]; while (j <= N) temp [k ++] = [J ++]; for (I = 0; I <K; I ++) A [first + I] = temp [I];} void mergesort (int A [], int first, int last, int temp []) {If (first <last) {int mid = (first + last)/2; mergesort (A, first, mid, temp); // ordered mergesort (A, Mid + 1, last, temp) on the left; // ordered mergearray (A, first, mid, last, temp); // merge two more ordered columns} bool mergesort (int A [], int N) {int * P = new int [N]; if (P = NULL) return false; mergesort (A, 0, N- 1, P); Return true;} int main () {const int maxn = 5005; int A [maxn], B [maxn]; int t; while (~ Scanf ("% d", & T) {for (INT I = 0; I <t; I ++) scanf ("% d ", & A [I]), B [I] = A [I]; g_ncount = 0; mergesort (A, T); // printf ("the reverse order number pair is: % d \ n ", g_ncount); int ans = g_ncount; int sum = g_ncount; For (INT I = 0; I <t; I ++) {// printf ("% d \ n", sum); sum = sum-(B [I]) + (t-1-B [I]); if (ANS> sum) ans = sum;} printf ("% d \ n", ANS);} return 0 ;}

HDU 1394 minimum inversion number [number of reverse orders]