分治与归并

题目链接：https://vjudge.net/contest/243680#problem/F

密码（hpuacm）

F - Ultra-QuickSort

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

标准的分治与归并板子题，这里用的是小白书上更腻害的方法哟！

Tis: 注意数组大小

#include <cstdio>
#include <vector>
using namespace std;

typedef long long ll;
vector<int> A;

ll countt(vector <int> &a)
{
    int n = a.size();
    if (n <= 1) return 0;
    ll cnt = 0;
    vector <int> b(a.begin(),a.begin() + n / 2);
    vector <int> c(a.begin() + n / 2, a.end());

    cnt += countt(b);//(1)
    cnt += countt(c);//(2)

    //(3)
    int ai = 0, bi = 0, ci = 0;
    while(ai < n)
    {
        if(bi < b.size() && (ci == c.size() || b[bi] <= c[ci]))
        {
            a[ai++] = b[bi++];
        }
        else
        {
            cnt += n / 2 - bi;
            a[ai++] = c[ci++];

        }
    }
    return cnt;
}

int main()
{
    int t,x;
    while(scanf("%d", &t), t)
    {
        A.clear();
        for(int i = 0; i < t; i++)
        {
            scanf("%d", &x);
            A.push_back(x);
        }
        printf("%lld\n", countt(A));
    }

    return 0;
}