1、http://blog.csdn.net/sdjzping/article/details/20381665
2、题目大意:
有n个数字，定义i<j并且a[i]>a[j]这样的一对数为逆序对，求这个数中有多少逆序对，
3、题目分析
看着题目很简单，但是n=65537,两重for循环找必超时，所以想到了用树状数组，但是交了好几遍都错了，runtime error,原因就是数组中的数字可以到达10^9，数组肯定要超内存，此题需要用到离散化，但是数字之间可能有重复，这一点没考虑到，有wrong 了好几遍，不过终于是改对了，。
4、题目：
180. Inversions
time limit per test: 0.25 sec.
memory limit per test: 4096 KB
memory limit per test: 4096 KB
input: standard
output: standard
output: standard
There are N integers (1<=N<=65537) A1, A2,.. AN (0<=Ai<=10^9). You need to find amount of such pairs (i, j) that 1<=i<j<=N and A[i]>A[j].
Input
The first line of the input contains the number N. The second line contains N numbers A1...AN.
Output
Write amount of such pairs.
Sample test(s)
Input
5 2 3 1 5 4
Output
3
5、AC代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 65540
#define ll long long
int n;
int c[N];
struct node
{int id;int v;
}a[N];
node b[N];
int cmp(node a,node b)
{return a.v<b.v;
}
int cmp1(node a,node b)
{return a.id<b.id;
}int lowbit(int i)
{return i&(-i);
}
void update(int x,int v)
{for(int i=x;i<=n;i+=lowbit(i)){c[i]+=v;}
}
ll getsum(int x)
{ll sum=0;//此处注意需要计算的是比自己小的，所以不包含自己for(int i=x-1;i>=1;i-=lowbit(i)){sum+=c[i];}return sum;
}
int main()
{while(scanf("%d",&n)!=EOF){memset(c,0,sizeof(c));for(int i=1;i<=n;i++){scanf("%d",&a[i].v);a[i].id=i;}sort(a+1,a+n+1,cmp);a[0].v=-1;int k=1;//注意离散化的时候考虑相等的情况for(int i=1;i<=n;i++){if(a[i].v!=a[i-1].v){b[i].id=a[i].id;b[i].v=k++;}else{b[i].id=a[i].id;b[i].v=b[i-1].v;}}sort(b+1,b+n+1,cmp1);ll sum=0;for(int i=n;i>=1;i--){sum+=getsum(b[i].v);update(b[i].v,1);}printf("%lld\n",sum);}return 0;
}