Nyoj 233 &&nyoj 322 Sort (tree-like array)

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Test instructions

Describe

You want to processe a sequence of n distinct integers by swapping II adjacent sequence elements until the sequence I s 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.

Input

The input consists of T number of test cases. (<0t<1000) Each case consists of the 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.

Output

For each case, the output of the minimum times need to sort it in ascending order on a single line.

Sample input

231 2 344 3 2 1

Sample output

06

idea: is to ask for a set of data in reverse order number, tree-like array method, do not explain:

Code:

#include <math.h> #include <queue> #include <deque> #include <vector> #include <stack># Include <stdio.h> #include <ctype.h> #include <string.h> #include <stdlib.h> #include < Iostream> #include <algorithm>using namespace std; #define LOWBIT (a) A&-a#define Max (A, b) a>b?a:b# Define Min (A, B) a>b?b:a#define mem (A, B) memset (A,b,sizeof (a)) int dir[4][2]= {{1,0},{-1,0},{0,1},{0,-1}};const Double EPS = 1e-6;const double Pi = ACOs ( -1.0); static const int inf= ~0u>>2;static const int MAXN =110;int h[100000        1],w[100],map[200];int m[10001], n;void Add (int i) {while (i<=1001) {m[i]++;    I+=lowbit (i);    }}int Sum (int i) {int res=0;        while (i>0) {res+=m[i];    I-=lowbit (i); } return res;    int main () {int t,temp;    scanf ("%d", &t);        while (t--) {scanf ("%d", &n);        memset (M,0,sizeof (m));        int ans=0; for (int i=1; i<=n; i++) {SCANF ("%d", &temp);            ADD (temp);        ans+= (I-sum (temp));    } printf ("%d\n", ans); } return 0;}

When you want-give up, think for why are you persist until now！

Nyoj 233 &&nyoj 322 Sort (tree-like array)