Problem Description
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.
For example, 1 2 3 5 4, we only need one operation : swap 5 and 4.
Input
The input consists of a number of test cases. 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.
Output
For each case, output the minimum times need to sort it
in ascending order on a single line.
Sample Input
3
1 2 3
4
4 3 2 1
Sample Output
0
6
为什么可以用逆序数呢，因为它要相邻两个交换，而逆序数，比如 4 3 2 1，比3大的1个，正好4，3交换，比2大的2个，4和3，2先和3换，再和4换。就算4和3先不进行交换，2与3交换，变成4 2 3 1，在3前面比3大的还是4，还是会通过2，4交换，然后再3，4交换，所以这题可以通过求逆序数解
AC代码：
1 #include<cstdio> 2 #include<cstring> 3 const int N = 100000+5; 4 int n; 5 int c[N],a[N]; 6 7 int lowbit(int x) { 8 return x&-x; 9 } 10 11 void add(int x, int a) { 12 while(x <= n) { 13 c[x] += a; 14 x += lowbit(x); 15 } 16 } 17 18 int sum(int x) { 19 int ans = 0; 20 while(x) { 21 ans += c[x]; 22 x -= lowbit(x); 23 } 24 return ans; 25 } 26 27 int main(){ 28 while(scanf("%d",&n)!=EOF) { 29 memset(c,0,sizeof(c)); 30 int ans=0; 31 for(int i = 1; i <= n; i++) { 32 scanf("%d",&a[i]); 33 add(a[i], 1); 34 ans += i-sum(a[i]); 35 } 36 printf("%d ",ans); 37 } 38 return 0; 39 }