逆序对的数量（Acwing）

 

1.首先要想到排序问题中的归并排序来解决此问题；

其次我们要看逆序数的定义是i<j&&a[i]>a[j] ；

下面就来模拟一下；

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#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
typedef long long LL;
int tmp[N];
LL merge_sort(int q[], int l, int r)
{
    if (l >= r) return 0;

    int mid = l + r >> 1;

    LL res = 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
        {
            res += mid - i + 1;
            tmp[k ++ ] = q[j ++ ];
        }
    while (i <= mid) tmp[k ++ ] = q[i ++ ];
    while (j <= r) tmp[k ++ ] = q[j ++ ];

    for (i = l, j = 0; i <= r; i ++, j ++ ) q[i] = tmp[j];

    return res;
}
int main(){
    int n;
    cin>>n;
    int q[N];
    for(int i=0;i<n;i++) scanf("%d",&q[i]);
    cout<<merge_sort(q,0,n-1);
    return 0;
}