2299 Ultra-QuickSort

Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536KTotal 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

0

5

4

3

1

2

3

0

Sample Output

6

0

Source

Waterloo local 2005.02.05

#include <iostream>#include <stdio.h>#include <stdlib.h>using namespace std;#define MAX 500001int n, a[MAX], t[MAX];long long sum;/* 归并 */void Merge(int l, int m, int r) {    /* p指向输出区间 */    int p = 0;    /* i、j指向2个输入区间 */    int i = l, j = m + 1;    /* 2个输入区间都不为空时 */    while(i <= m && j <= r) {        /* 取关键字小的记录转移至输出区间 */        if (a[i] > a[j]) {            t[p++] = a[j++];            /* a[i]后面的数字对于a[j]都是逆序的 */            sum += m - i + 1;        }else {            t[p++] = a[i++];        }    }    /* 将非空的输入区间转移至输出区间 */    while(i <= m) t[p++] = a[i++];    while(j <= r) t[p++] = a[j++];    /* 归并完成后将结果复制到原输入数组 */    for (i = 0; i < p; i++){        a[l + i] = t[i];    }}/* 归并排序 */void MergeSort(int l, int r) {    int m;    if (l < r) {        /* 将长度为n的输入序列分成两个长度为n/2的子序列 */        m = (l + r) / 2;        /* 对两个子序列分别进行归并排序 */        MergeSort(l, m);        MergeSort(m + 1, r);        /* 将2个排好的子序列合并成最终有序序列 */        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;}