【归并排序】Acwing788.逆序对的数量

Acwing788.逆序对的数量

题解

暴力做法：对每个字符串前或后二选一进行统计，复杂度O(n2)超时
归并排序：由于归并二分的特性，能够保证每个数在没排序前都能统计到（会分到长度为2的序列），同时排序完左右后并不影响右边对于左边的逆序对数，或左边对于右边的逆序对数，时间复杂度O(nlogn)

#include <iostream>
#include <cstdio>

using namespace std;

typedef long long LL;

const int N = 1e5 + 10;

int q[N], tmp[N];


LL merge(int l, int r)
{
    if(l == r) return 0;
    int mid = l + r >> 1;
    LL res = merge(l, mid) + merge(mid + 1, r);
    int i = l, j = mid + 1, k = l;
    while(i <= mid && j <= r)
    {
        if(q[i] > q[j])
        {
            res += mid - i + 1;
            tmp[k ++ ] = q[j ++ ];
        }
        else tmp[k ++ ] = q[i ++ ];
    }
    while(i <= mid) tmp[k ++ ] = q[i ++ ];
    while(j <= r) tmp[k ++ ] = q[j ++ ];
    for(int x = l; x < k; ++x) q[x] = tmp[x];
    return res;
}

int main()
{
    int n;
    scanf("%d",&n);
    for(int i = 0; i < n; ++i)
        scanf("%d",&q[i]);
    cout << merge(0, n - 1) << endl;
    return 0;
}