O(nlogn)求逆序数对的个数

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<cmath>
using namespace std;
const int maxn=1e5+5;
int is1[maxn],is2[maxn];// is1为原数组，is2为临时数组，n为个人定义的长度

long long merge(int low,int mid,int high)
{
    int i=low,j=mid+1,k=low;
    long long count=0;
    while(i<=mid&&j<=high)
        if(is1[i]<=is1[j])// 此处为稳定排序的关键，不能用小于
            is2[k++]=is1[i++];
        else
        {
            is2[k++]=is1[j++];
            count+=j-k;// 每当后段的数组元素提前时，记录提前的距离
        }
    while(i<=mid)
        is2[k++]=is1[i++];
    while(j<=high)
        is2[k++]=is1[j++];
    for(i=low;i<=high;i++)// 写回原数组
        is1[i]=is2[i];
    return count;
}
long long mergeSort(int a,int b)// 下标，例如数组int is[5]，全部排序的调用为mergeSort(0,4)
{
    if(a<b)
    {
        int mid=(a+b)/2;
        long long count=0;
        count+=mergeSort(a,mid);
        count+=mergeSort(mid+1,b);
        count+=merge(a,mid,b);
        return count;
    }
    return 0;
}
int main()
{
    int n,x,y;
    while(~scanf("%d%d%d",&n,&x,&y))
    {
        int tmp=min(x,y);
        for(int i=1;i<=n;i++)
            scanf("%d",&is1[i]);
        printf("%lld
",tmp*mergeSort(1,n));
    }
    return 0;
}