在数组中的两个数字，如果前面一个数字大于后面的数字，则这两个数字组成一个逆序对。输入一个数组,求出这个数组中的逆序对的总数P。并将P对1000000007取模的结果输出。 即输出P%1000000007

include "stdafx.h"

#include<iostream>
#include<vector>
#include <algorithm>  
#include<iomanip>
#include<string>

using namespace std;
class Solution {
public:
	long long nums = 0;
	int InversePairs(vector<int> data) {
		vector<int> temp=data;
		MSort(data, temp, 0, data.size()-1);
	//	cout << "nums:" << nums << endl;
		return nums % 1000000007;
	}

	void Merge(vector<int> &data, vector<int> &temp, int begin, int mid, int end)
	{
		int start = begin;//记录开始位置
		int left = begin;
		int right = mid + 1;
		while (left<=mid && right <=end)
		{
			if (data[left] < data[right])
			{
				temp[begin++] = data[left++];
			}
			else
			{
				temp[begin++] = data[right++];
				nums += mid - left + 1;
			}
		}
		while (left<=mid)
		{
			temp[begin++] = data[left++];
		}
		while (right <= end)
		{
			temp[begin++] = data[right++];
		}
		while (start<=end)
		{
			data[start] = temp[start];
			start++;
		}
	}

	void MSort(vector<int> &data, vector<int> &temp, int begin,int end)
	{
		int mid;
		if (begin < end)
		{
			mid = (begin + end) / 2;
			MSort(data, temp, begin, mid);
			MSort(data, temp, mid+1,end);
			Merge(data, temp, begin, mid, end);
		}
	}
};

int main()
{

	vector<int> data = { 1,2,3,4,5,6,7,0 };
	Solution so;
	so.InversePairs(data);
	return 0;
}