求逆序对的解法

一、归并排序解法

#include <bits/stdc++.h>
using namespace std;
int n,a[2000001],temp[2000001];
long long ans;
void msort(int L,int R)
{
    int mid = (L + R) / 2;
    if(R > L)
    {
        msort(L,mid);
        msort(mid+1,R);
        int i = L,k = L,j = mid + 1;
        while(i<=mid && j<=R)
        {
            if(a[i] > a[j])
            {
                temp[k++] = a[j++];
                ans += mid - i + 1;
            }
            else
                temp[k++] = a[i++];
        }
        while(i<=mid) temp[k++] = a[i++];
        while(j<=R) temp[k++] = a[j++];
        for(int x = L;x<=R;x++) a[x] = temp[x];
    }
    
}
int main()
{
    cin>>n;
    for(int i = 0;i<n;i++)
        cin>>a[i];
        
    msort(0,n-1);
    
    cout<<ans<<endl;
        
    return 0;
} 

　　这是求小的个数的树状数组做法：

#include <cstdio>
#include <cstring>
const int maxn = 100010;
#define lowbit(i) ((i)&(-i))
int c[maxn];

void update(int x, int v)
{
    for(int i = x;i < maxn; i += lowbit(i))
        c[i] += v;
}

int getSum(int x)
{
    int sum = 0;
    for(int i = x;i>0;i -= lowbit(i))
        sum += c[i];
    return sum;
}

int main()
{
    int n,x,ans = 0;
    scanf("%d",&n);
    memset(c,0,sizeof(c));
    for(int i = 0;i<n;i++)
    {
        scanf("%d",&x);
        update(x,1);
        printf("%d\n",getSum(x-1));
    }
}

树状数组正解解法：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#include<string>
#include<cstring>
#define int long long int
#define lowbit(x) x & -x
const int N=5e5+10;
using namespace std;
inline int read() {
    char c = getchar();
    int x = 0, f = 1;
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int ans;
int tree[N],n,a[N],b[N];
void add(int x)
{
    while(x<=n)
    {
        tree[x]++;
        x+=lowbit(x);
    }
}
int query(int x)
{
    int ans=0;
    while(x>0)
    {
        ans+=tree[x];
        x-=lowbit(x);   
    }
    return ans;
}
signed main()
{
    n=read();
    for(int i=1;i<=n;++i) a[i]=read(),b[i]=a[i];
    sort(a+1,a+1+n);
    int len=unique(a+1,a+1+n)-a-1;
    for(int i=1;i<=n;++i)
    {
        int p=lower_bound(a+1,a+1+len,b[i])-a;
        add(p);
        ans+=i-query(p);
    }
    cout<<ans;
    return 0;
}