POJ 2299：Ultra-QuickSort

Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536KTotal Submissions: 39397 Accepted: 14204

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

59105431230

Sample Output

60

归并排序。

另外，此题有一坑就是结果会超int32；

具体可以参考：点击打开链接

我写的代码如下：

#include<cstdio>#include<stdlib.h>#include<cstring>#include<algorithm>#include<iostream>using namespace std;const int M = 500000 + 5;int n, A[M], T[M], i;long long merge_sort(int l, int r, int *A){    if (r - l < 1) return 0;    int mid = (l + r) / 2;    long long ans = merge_sort(l, mid, A) + merge_sort(mid + 1, r, A);    i = l;    int p = l, q = mid + 1;    while (p <= mid && q <= r)    {        if(A[p] <= A[q])            T[i++] = A[p++];        else        {            ans += (mid + 1 - p);            T[i++] = A[q++];        }    }    while (p <= mid) T[i++] = A[p++];    while (q <= r) T[i++] = A[q++];    for (int j = l; j <= r; j++)        A[j] = T[j];    return ans;}int main(){    int n;    while(scanf("%d", &n) && n)    {        for(int j=0; j<n; j++)            scanf("%d", &A[j]);       printf("%lld\n", merge_sort(0, n - 1, A));    }    return 0;}