2299 Ultra-QuickSort

Ultra-QuickSort

Time Limit: 7000MS	Memory Limit: 65536K
Total Submissions: 24055	Accepted: 8603

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 

9 1 0 5 4 ,

Ultra-QuickSort produces the output 

0 1 4 5 9 .

Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given 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

5      

9      

1      

5      

4      

3      

1      

2      

3      

Sample Output

6      

Source

Waterloo local 2005.02.05

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
using namespace std;
#define MAX 500001

int n, a[MAX], t[MAX];
long long sum;

/* 歸并 */
void Merge(int l, int m, int r) {
    /* p指向輸出區間 */
    int p = 0;
    /* i、j指向2個輸入區間 */
    int i = l, j = m + 1;
    /* 2個輸入區間都不為空時 */
    while(i <= m && j <= r) {
        /* 取關鍵字小的記錄轉移至輸出區間 */
        if (a[i] > a[j]) {
            t[p++] = a[j++];
            /* a[i]後面的數字對于a[j]都是逆序的 */
            sum += m - i + 1;
        }else {
            t[p++] = a[i++];
        }
    }
    /* 将非空的輸入區間轉移至輸出區間 */
    while(i <= m) t[p++] = a[i++];
    while(j <= r) t[p++] = a[j++];
    /* 歸并完成後将結果複制到原輸入數組 */
    for (i = 0; i < p; i++){
        a[l + i] = t[i];
    }
}

/* 歸并排序 */
void MergeSort(int l, int r) {
    int m;
    if (l < r) {
        /* 将長度為n的輸入序列分成兩個長度為n/2的子序列 */
        m = (l + r) / 2;
        /* 對兩個子序列分别進行歸并排序 */
        MergeSort(l, m);
        MergeSort(m + 1, r);
        /* 将2個排好的子序列合并成最終有序序列 */
        Merge(l, m, r);
    }
}

int main() {
    int i;
    while(cin >> n) {
        if (n == 0) break;
        sum=0;
        for(i = 0; i < n; i++) {
            scanf("%d", &a[i]);
        }
        MergeSort(0, n - 1);
        printf("%I64d\n", sum);
    }
    return 0;
}