poj2299(求逆序对，归并排序)

  归并排序求逆序对了，只要是逆序对的，肯定就会有交换的操作了，然后归并排序的时候，此时新数组放右边的数字时，计算逆序对个数（就是还未放入的个数），然后就可以得到答案了。

  

#include "stdio.h"
#include "string.h"
#include "math.h"
#include <string>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <algorithm>
#include <iostream>
using namespace std;

#define MAXM 1
#define MAXN 1
#define max(a,b) a > b ? a : b
#define min(a,b) a < b ? a : b
#define Mem(a,b) memset(a,b,sizeof(a))
int Mod = 1000000007;
double pi = acos(-1.0);
double eps = 1e-6;

typedef struct{
	int f,t,w,next;
}Edge;
Edge edge[MAXM];
int head[MAXN];
int kNum;

void addEdge(int f, int t, int w)
{
	edge[kNum].f = f;
	edge[kNum].t = t;
	edge[kNum].w = w;
	edge[kNum].next = head[f];
	head[f] = kNum ++;
}

int dp[500005];
int N;
int T[500005];
__int64 ans;

void Merge(int s, int t)
{
	if( s >= t )
		return ;
	int mid = (s + t) >> 1; 
	Merge(s, mid);
	Merge(mid+1, t);
	int p = mid + 1, q = t + 1;
	int i = s, j = mid + 1, k = s;


	while( i < p && j < q ){
		if( dp[i] <= dp[j] ){
			T[k++] = dp[i++];
		}
		else{
			T[k++] = dp[j++];
			//计数
			ans += ( p - i );
		}
	}

	if( i == p ){
		for( ; j < q; j ++){
			T[k++] = dp[j];
		}
	}
	else{
		for(; i < p; i ++){
			T[k++] = dp[i];
		}
	}

	for(i = s; i <= t; i ++){
		dp[i] = T[i];
	}

}

void solve(){
	for(int i = 0; i < N; i ++){
		scanf("%d",&dp[i]);
	}
	ans = 0;
	Merge(0,N - 1);

	printf("%I64d\n",ans);
}

int main()
{
//	freopen("d:\\test.txt", "r", stdin);
	
	while(cin>>N){
		if( N == 0 )
			break;
		solve();
	}
	

	return 0;
}