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nyoj322Sort

时间限制：1000 ms | 内存限制：65535 KB

难度：4

描述

You want to processe a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is 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.

输入

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

输出

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

样例输入

2

3

1 2 3

4

4 3 2 1

样例输出

0

6

//归并排序//http://blog.csdn.net/yuehailin/article/details/68961304 #include
   
    #include
    
     using namespace std;void mergeArray(int number[], int first, int mid, int last, int temp[]);void mergeSort(int number[], int first, int last, int temp[]);int count;int main() {  int t, n, number[1005], temp[1005];  scanf("%d", &t);  while (t--) {    count = 0;    scanf("%d", &n);    for (int i = 0; i < n; i++) scanf("%d", &number[i]);    mergeSort(number, 0, n-1, temp);//    for (int i = 0; i < n; i++) printf("%d ", number[i]);//    printf("\n");    printf("%d\n", count);  }  return 0;}void mergeArray(int number[], int first, int mid, int last, int temp[]) {  int i = first, j = mid + 1;  int position = 0;  while (i <= mid && j <= last) {    if (number[i] <= number[j]) {      temp[position++] = number[i++];    }else {      count += mid - i + 1;      temp[position++] = number[j++];    }  }  while (i <= mid) temp[position++] = number[i++];  while (j <= last) temp[position++] = number[j++];  for (i = 0; i < position; i++) number[first+i] = temp[i];}void mergeSort(int number[], int first, int last, int temp[]) {  if (first < last) {    int mid = (first + last) / 2;    mergeSort(number, first, mid, temp);    mergeSort(number, mid+1, last, temp);    mergeArray(number, first, mid, last, temp);  }}
    
   

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