Petr and Permutations CodeForces - 987E

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交换次数的奇偶性和逆序对总数的奇偶性是一致的 当结论记住

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define ll long long

struct node
{
    int l;
    int r;
    ll val;
};

node tree[4000010];
ll pre[1000010];
int num[1000010];
int n;

void pushup(int cur)
{
    tree[cur].val=tree[cur*2].val+tree[cur*2+1].val;
    return;
}

ll query(int pl,int pr,int cur)
{
    ll res;
    if(pl<=tree[cur].l&&tree[cur].r<=pr)
    {
        return tree[cur].val;
    }
    res=0;
    if(pl<=tree[cur*2].r) res+=query(pl,pr,cur*2);
    if(pr>=tree[cur*2+1].l) res+=query(pl,pr,cur*2+1);
    return res;
}

void update(int tar,int cur)
{
    if(tree[cur].l==tree[cur].r)
    {
        tree[cur].val=1;
        return;
    }
    if(tar<=tree[cur*2].r) update(tar,cur*2);
    else update(tar,cur*2+1);
    pushup(cur);
    return;
}

void build(int l,int r,int cur)
{
    int m;
    tree[cur].l=l;
    tree[cur].r=r;
    tree[cur].val=0;
    if(l==r) return;
    m=(l+r)/2;
    build(l,m,cur*2);
    build(m+1,r,cur*2+1);
    pushup(cur);
    return;
}

int main()
{
    ll ans;
    int i;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        scanf("%d",&num[i]);
    }
    build(1,n,1);
    for(i=1;i<=n;i++)
    {
        if(num[i]<n)
        {
            pre[i]+=query(num[i]+1,n,1);
        }
        update(num[i],1);
    }
    build(1,n,1);
    for(i=n;i>=1;i--)
    {
        if(num[i]>1)
        {
            pre[i]+=query(1,num[i]-1,1);
        }
        update(num[i],1);
    }
    ans=0;
    for(i=1;i<=n;i++)
    {
        ans+=pre[i];
    }
    ans/=2;
    if(ans%2==(3*n)%2) printf("Petr\n");
    else printf("Um_nik\n");
    return 0;
}