AcWing 788. 逆序对的数量
原题链接
简单
作者:
因是因非
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2022-02-24 21:43:05
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所有人可见
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阅读 155
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e7;
ll ans;
int tmp[N];
int a[N];
void gui(int q[], int l, int r)
{
if(l >= r) return ;
int mid = l + r >> 1;
gui(q, l, mid);
gui(q, mid + 1, r);
int i = l, j = mid + 1, k = 1;
while( i <= mid && j <= r)
{
if(q[i] <= q[j]) tmp[k ++ ] = q[i ++ ];
else
{
ans += mid - i + 1;
tmp[k ++ ] = q[j ++ ];
}
}
while(i <= mid) tmp[k ++ ] = q[i ++ ];
while(j <= r) tmp[k ++ ] = q[j ++ ];
for(int i = l, j = 1; i <= r; i ++ , j ++ )
q[i] = tmp[j];
}
int main()
{
int n;
cin >> n;
for(int i = 1; i <= n; i ++ )
scanf("%d", &a[i]);
gui(a, 1, n);
cout << ans;
}