Ultra-QuickSort
| Time Limit: 7000MS | Memory Limit: 65536K | |
| Total Submissions: 32009 | Accepted: 11395 |
Description
In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence Input
The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.
Output
For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.
Sample Input
59105431230
Sample Output
60 代码 ：
#include#include #include using namespace std ;#define M 500010int xt[M*2] , n ;int a[M] ;struct node{ int w , id ; } qe[M] ;// 用结构体储存 原始数据 id 是数据原始的位置int cmp( node a , node b ){ return a.w < b.w ;}int lowbit( int x ){ return x & ( -x ) ;}void update( int x ){ while( x <= n ){ xt[x] += 1 ; x += lowbit(x) ; }}int sum( int x ){ int s = 0 ; while( x > 0 ){ s += xt[x] ; x -= lowbit(x) ; } return s ;}int main(){ int i ; long long mun ; while( cin >> n ){ if( n == 0 ) break ; mun = 0 ; memset( xt , 0 , sizeof(xt) ) ; for( i = 1; i <= n ;i++){ scanf( "%d" , &qe[i].w ) ; qe[i].id = i ; } sort( qe + 1 , qe + 1 + n , cmp ) ; for( i = 1 ; i <= n ; i++){ a[qe[i].id] = i ; // 对数据进行离散化 例如 原来是 9 0 1 2 处理后是 4 1 2 3 } for( i = 1 ; i <= n ; i++){ update(a[i] ) ; mun += ( i - sum(a[i] - 1 ) - 1 ) ; } cout << mun << endl ; }}