（归并排序）AcWing 788. 逆序对的数量

目录

题目链接

一些话

流程

套路

ac代码

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题目链接

788. 逆序对的数量 – AcWing题库

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一些话

//注意cnt要开long long，最大是n方

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流程

在原归并排序的这部分做修改，else时加一个cnt += mid – i + 1即可

    while(i <= mid && j <= r) {
        if(q[i] <= q[j]) tmp[k++] = q[i++];
        else tmp[k++] = q[j++],cnt += mid – i + 1;
    }

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套路

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ac代码

[//]: # (打卡模板，上面预览按钮可以展示预览效果 ^^)
```
//注意cnt要开long long，最大是n方
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;
const int N = 1e5 + 10;
int tmp[N],q[N];
int n;
long long cnt = 0;
void merge_sort(int q[],int l,int r){
    if(l >= r) return ;
    
    int mid = l + r >> 1;
    
    merge_sort(q,l,mid);
    merge_sort(q,mid+1,r);
    int k = 0,i = l,j = mid+1;
    while(i <= mid && j <= r) {
        if(q[i] <= q[j]) tmp[k++] = q[i++];
        else tmp[k++] = q[j++],cnt += mid - i + 1;
    }
    while(i <= mid) tmp[k++] = q[i++];
    while(j <= r) tmp[k++] = q[j++];
    //这里不熟
    for(int i = l,j = 0;i <= r;i++,j++){
        q[i] = tmp[j];
    }
}
int main(){
    
    cin >> n;
    for(int i = 0;i < n;i++){
        scanf("%d",&q[i]);
    }
    merge_sort(q,0,n-1);
    cout << cnt << endl;
}
```

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