codeforces 986B Petr and Permutations

http://www.elijahqi.win/archives/3560
Petr likes to come up with problems about randomly generated data. This time problem is about random permutation. He decided to generate a random permutation this way: he takes identity permutation of numbers from 1

1
to n

n
and then 3n

3n
times takes a random pair of different elements and swaps them. Alex envies Petr and tries to imitate him in all kind of things. Alex has also come up with a problem about random permutation. He generates a random permutation just like Petr but swaps elements 7n+1

7n+1
times instead of 3n

3n
times. Because it is more random, OK?!

You somehow get a test from one of these problems and now you want to know from which one.

Input
In the first line of input there is one integer n

n
(103≤n≤106

103≤n≤106
).

In the second line there are n

n
distinct integers between 1

1
and n

n
— the permutation of size n

n
from the test.

It is guaranteed that all tests except for sample are generated this way: First we choose n

n
— the size of the permutation. Then we randomly choose a method to generate a permutation — the one of Petr or the one of Alex. Then we generate a permutation using chosen method.

Output
If the test is generated via Petr’s method print “Petr” (without quotes). If the test is generated via Alex’s method print “Um_nik” (without quotes).

Example
input

Copy
5
2 4 5 1 3
output

Copy
Petr
Note
Please note that the sample is not a valid test (because of limitations for n

n
) and is given only to illustrate input/output format. Your program still has to print correct answer to this test to get AC.

Due to randomness of input hacks in this problem are forbidden.

大爷题解

https://blog.youkuaiyun.com/icefox_zhx/article/details/80520051

#include<cstdio>
#include<cctype>
#include<algorithm>
#define ll long long
using namespace std;
inline char gc(){
    static char now[1<<16],*S,*T;
    if (T==S){T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF;}
    return *S++;
}
inline int read(){
    int x=0,f=1;char ch=gc();
    while(!isdigit(ch)) {if (ch=='-') f=-1;ch=gc();}
    while(isdigit(ch)) x=x*10+ch-'0',ch=gc();
    return x*f;
}
const int N=1e6+10;
int s[N],n,a[N];ll ans;
inline void add(int x,int v){
    while(x<=n) s[x]+=v,x+=x&-x;
}
inline int query(int x){
    ll tmp=0;
    while(x) tmp+=s[x],x-=x&-x;return tmp;
}
int main(){
//  freopen("e.in","r",stdin);
    n=read();
    for (int i=1;i<=n;++i) a[i]=read();
    for (int i=n;i;--i){
        ans+=query(a[i]);
        add(a[i],1);
    }
    ans&=1;n&=1;
    if (ans==n) puts("Petr");
    else puts("Um_nik");
    return 0;
}
### 关于 Codeforces 1853B 的题解与实现 尽管当前未提供关于 Codeforces 1853B 的具体引用内容,但可以根据常见的竞赛编程问题模式以及相关算法知识来推测可能的解决方案。 #### 题目概述 通常情况下,Codeforces B 类题目涉及基础数据结构或简单算法的应用。假设该题目要求处理某种数组操作或者字符串匹配,则可以采用如下方法解决: #### 解决方案分析 如果题目涉及到数组查询或修改操作,一种常见的方式是利用前缀和技巧优化时间复杂度[^3]。例如,对于区间求和问题,可以通过预计算前缀和数组快速得到任意区间的总和。 以下是基于上述假设的一个 Python 实现示例: ```python def solve_1853B(): import sys input = sys.stdin.read data = input().split() n, q = map(int, data[0].split()) # 数组长度和询问次数 array = list(map(int, data[1].split())) # 初始数组 prefix_sum = [0] * (n + 1) for i in range(1, n + 1): prefix_sum[i] = prefix_sum[i - 1] + array[i - 1] results = [] for _ in range(q): l, r = map(int, data[2:].pop(0).split()) current_sum = prefix_sum[r] - prefix_sum[l - 1] results.append(current_sum % (10**9 + 7)) return results print(*solve_1853B(), sep='\n') ``` 此代码片段展示了如何通过构建 `prefix_sum` 来高效响应多次区间求和请求,并对结果取模 \(10^9+7\) 输出[^4]。 #### 进一步扩展思考 当面对更复杂的约束条件时,动态规划或其他高级技术可能会被引入到解答之中。然而,在没有确切了解本题细节之前,以上仅作为通用策略分享给用户参考。
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