题解:直接找出来没个数交换的次数就好了。。。。没个数交换的次数是这个数左边比它大的+右边比它小的,注意是LL 啊 !!!错了半天;还有重复值的处理;
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define SD(x,y) scanf("%lf%lf",&x,&y)
#define P_ printf(" ")
typedef long long LL;
const int MAXN=1000010;
int tree[MAXN];
int b[MAXN],a[MAXN];
LL dp[MAXN];
int lowbit(int x){return x&(-x);}
void insert(int x){
	while(x<MAXN){
		tree[x]++;
		x+=lowbit(x);
	}
}
int Sum(int x){
	int sum=0;
	while(x>0){
		sum+=tree[x];
		x-=lowbit(x);
	}
	return sum;
}
int main(){
	int n;
	SI(n);
	LL ans=0;
	dp[0]=0;
	for(int i=1;i<=100000;i++)dp[i]=dp[i-1]+i;
	mem(b,0);mem(tree,0);
	for(int i=0;i<n;i++){
		SI(a[i]);
		b[i]=i-Sum(a[i]+1);
		insert(a[i]+1);
	}
//	for(int i=0;i<n;i++)PI(b[i]);puts("");
	mem(tree,0);
	for(int i=n-1;i>=0;i--){
		insert(a[i]+1);
		b[i]+=Sum(a[i]);
	}
//	for(int i=0;i<n;i++)PI(b[i]);puts("");
	ans=0;
	for(int i=0;i<n;i++)ans+=dp[b[i]];
	printf("%lld\n",ans);
	return 0;
}
原文:http://www.cnblogs.com/handsomecui/p/5122776.html