【结论】升序逆序对

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求法：
1、归并排序

#include<iostream>#include<cstdio> using namespace std;int a[10003],a1[10003];int ans=0;void merge(int l,int mid,int r){    int i=l,k=l,j=mid+1;    while(i<=mid&&j<=r)    {        if(a[i]<=a[j]) a1[k++]=a[i++];// <        else         {            a1[k++]=a[j++];            ans+=mid-i+1;        }    }    while(i<=mid) a1[k++]=a[i++];    while(j<=r) a1[k++]=a[j++];    for(i=l;i<=r;i++) a[i]=a1[i]; }void mergesort(int l,int r){    if(l<r)    {        int mid=(l+r)>>1;        mergesort(l,mid);        mergesort(mid+1,r);        merge(l,mid,r);    } } int main(){    int n;scanf("%d",&n);    for(int i=1;i<=n;i++) scanf("%d",&a[i]);    mergesort(1,n);    printf("%d",ans); }

2、树状数组
(1)、当数字范围大于0且不大时，直接求

#include<cstdio>#include<iostream>using namespace std;const int need=20003;//数字范围为20000； int c[need];int ans=0;inline int lowbit(int s){return s&-s;}int getsum(int x){    int cnt=0;    for(int i=x;i>0;i-=lowbit(i)) cnt+=c[i];    return cnt;} void modify(int x){    for(int i=x;i<need;i+=lowbit(i)) c[i]++; } int main(){    int n;scanf("%d",&n);    for(int i=1,d;i<=n;i++)    {        scanf("%d",&d);        ans+=getsum(need-1)-getsum(d);// getsum(d-1)        modify(d);    }    printf("%d",ans);}

(2)、离散化

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