全部RE...

秦冇甪饕 @ 2022-06-20 19:14:59

#include <bits/stdc++.h>
using namespace std;
int a[1000]={},temp[1000]={},sum=0;
void msort(int l,int r)
{
    if(l>=r) return;
    int mid=(l+r)/2;
    msort(l,mid);
    msort(mid+1,r);
    int i=l,j=mid+1,k=l;
    while(i<=mid&&j<=r)
    {
        if(a[i]<a[j]) temp[k++]=a[i++];
        else 
        {
            temp[k++]=a[j++];
            sum+=(mid-i+1);
        }
    }
    while(i<=mid) temp[k++]=a[i++];
    while(j<=r) temp[k++]=a[j++];
    for(int o=l;o<=r;o++) a[o]=temp[o];
}
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
    } 
    msort(1,n);
    cout<<sum;
    return 0;
}

by MujicaSaki @ 2022-06-20 19:15:56

@秦冇甪饕 正确的

by MujicaSaki @ 2022-06-20 19:16:13

@秦冇甪饕 数组开小了

by _DeadPig_ @ 2022-06-20 19:18:27

对于所有数据，n \leq 5 \times 10^5

by Y2y7m @ 2022-06-20 19:19:57

@秦冇甪饕 做题不看数据那你会觉得你能AKNOI

by Y2y7m @ 2022-06-20 19:21:47

@秦冇甪饕 long long 没开！

by 秦冇甪饕 @ 2022-06-20 19:24:06

谢谢啊！虽然口气比较强硬...

by 秦冇甪饕 @ 2022-06-20 19:25:07

#include <bits/stdc++.h>
using namespace std;
long long a[500003]={},temp[500003]={},sum=0;
void msort(long long l,long long r)
{
    if(l>=r) return;
    long long mid=(l+r)/2;
    msort(l,mid);
    msort(mid+1,r);
    long long i=l,j=mid+1,k=l;
    while(i<=mid&&j<=r)
    {
        if(a[i]<a[j]) temp[k++]=a[i++];
        else 
        {
            temp[k++]=a[j++];
            sum+=(mid-i+1);
        }
    }
    while(i<=mid) temp[k++]=a[i++];
    while(j<=r) temp[k++]=a[j++];
    for(long long o=l;o<=r;o++) a[o]=temp[o];
}
int main()
{
    long long n;
    cin>>n;
    for(long long i=1;i<=n;i++)
    {
        cin>>a[i];
    } 
    msort(1,n);
    cout<<sum;
    return 0;
}

这样？