POJ - 2299 Ultra-QuickSort 【归并排序求逆序对】

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

#include<iostream>
using namespace std;
#include<cstring>
#include<cstdio>
#include<cstdlib>
long long cs = 0;
void merge(long long source[], long long temp[], long long l, long long mid, long long r) {
    long long i = l, j = mid + 1;
    long long point = l;
    while (i < mid + 1 && j < r + 1) {
        if (source[i] < source[j])
            temp[point++] = source[i++];
        else
            temp[point++] = source[j++],cs+=mid+1-i;
    }
    while (i < mid + 1) 
        temp[point++] = source[i++];
    while (j < r + 1) 
        temp[point++] = source[j++];
    for (i = l; i <= r; i++) 
        source[i] = temp[i];
}

void mergesort(long long source[], long long temp[], long long l, long long r) {
    if (l >= r) return;
    long long mid = (l + r) / 2;
    mergesort(source, temp, l, mid);
    mergesort(source, temp, mid + 1, r);
    merge(source, temp, l, mid, r);
}
long long a[510000];
long long t[510000];
int main(){
    long long n;
    while(scanf("%lld",&n)!=EOF&&n){
    for(long long i=0;i<n;i++)
        scanf("%lld",&a[i]);
    cs = 0;
    mergesort(a,t,0,n-1);
    cout<<cs<<endl;
    }
    return 0;
}