【CodeForces - 869】 A B C E 【组合数打表】

A The Artful Expedient
Rock… Paper!

After Karen have found the deterministic winning (losing?) strategy for rock-paper-scissors, her brother, Koyomi, comes up with a new game as a substitute. The game works as follows.

A positive integer n is decided first. Both Koyomi and Karen independently choose n distinct positive integers, denoted by x1, x2, …, xn and y1, y2, …, yn respectively. They reveal their sequences, and repeat until all of 2n integers become distinct, which is the only final state to be kept and considered.

Then they count the number of ordered pairs (i, j) (1 ≤ i, j ≤ n) such that the value xi xor yj equals to one of the 2n integers. Here xor means the bitwise exclusive or operation on two integers, and is denoted by operators ^ and/or xor in most programming languages.

Karen claims a win if the number of such pairs is even, and Koyomi does otherwise. And you’re here to help determine the winner of their latest game.

Input
The first line of input contains a positive integer n (1 ≤ n ≤ 2 000) — the length of both sequences.

The second line contains n space-separated integers x1, x2, …, xn (1 ≤ xi ≤ 2·106) — the integers finally chosen by Koyomi.

The third line contains n space-separated integers y1, y2, …, yn (1 ≤ yi ≤ 2·106) — the integers finally chosen by Karen.

Input guarantees that the given 2n integers are pairwise distinct, that is, no pair (i, j) (1 ≤ i, j ≤ n) exists such that one of the following holds: xi = yj; i ≠ j and xi = xj; i ≠ j and yi = yj.

Output
Output one line — the name of the winner, that is, “Koyomi” or “Karen” (without quotes). Please be aware of the capitalization.

Example
Input
3
1 2 3
4 5 6
Output
Karen
Input
5
2 4 6 8 10
9 7 5 3 1
Output
Karen
Note
In the first example, there are 6 pairs satisfying the constraint: (1, 1), (1, 2), (2, 1), (2, 3), (3, 2) and (3, 3). Thus, Karen wins since 6 is an even number.

In the second example, there are 16 such pairs, and Karen wins again.
题意两个 n个不同数的集合 A,B ，对于每对 < i j > ，A[i]^B[i] 的值如果在两个集合中任意一个，则cnt++，如果cnt为奇数输出 Koyomi ，偶数就输出 Karen 。

分析：数据很小，暴力就可以。
一开始用的 map，超时间，改为数组把，又数组越界了，两个在1e6 范围内的数字相异或值肯定是要大于1e6的。开大点就好，就是让1e6的二进制的数位上全变为1，应该就差不多。

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int>pii;
#define first fi
#define second se
#define  LL long long
#define fread() freopen("in.txt","r",stdin)
#define fwrite() freopen("out.txt","w",stdout)
#define CLOSE() ios_base::sync_with_stdio(false)const int MAXN = 2e3+10;
const int MAXM = 1e7;
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;int arr[MAXN];
int brr[MAXN];
bool vis[MAXM]={0} ;
int main(){CLOSE();
//  fread();
//  fwrite();int n;scanf("%d",&n);for(int i=1;i<=n;i++) {scanf("%d",&arr[i]);vis[arr[i]]=1;}for(int i=1;i<=n;i++) {scanf("%d",&brr[i]);vis[brr[i]] =1;}int cnt=0;for(int i=1;i<=n;i++){for(int j=1;j<=n;j++) {int c=arr[i]^brr[j];if(vis[c]) cnt++;}}if(cnt&1) puts("Koyomi");else puts("Karen");return 0;
}

B The Eternal Immortality

Even if the world is full of counterfeits, I still regard it as wonderful.

Pile up herbs and incense, and arise again from the flames and ashes of its predecessor — as is known to many, the phoenix does it like this.

The phoenix has a rather long lifespan, and reincarnates itself once every a! years. Here a! denotes the factorial of integer a, that is, a! = 1 × 2 × … × a. Specifically, 0! = 1.

Koyomi doesn’t care much about this, but before he gets into another mess with oddities, he is interested in the number of times the phoenix will reincarnate in a timespan of b! years, that is, . Note that when b ≥ a this value is always integer.

As the answer can be quite large, it would be enough for Koyomi just to know the last digit of the answer in decimal representation. And you’re here to provide Koyomi with this knowledge.

Input
The first and only line of input contains two space-separated integers a and b (0 ≤ a ≤ b ≤ 1018).

Output
Output one line containing a single decimal digit — the last digit of the value that interests Koyomi.

Example
Input
2 4
Output
2
Input
0 10
Output
0
Input
107 109
Output
2
Note
In the first example, the last digit of is 2;

In the second example, the last digit of is 0;

In the third example, the last digit of is 2.

分析：其实只要当便利过程中又0生成我们就可以停止了，之后都会是0，因为此时已经有因子2和5了。
可以想一下，要判断一个数n的阶乘最后一位是不是有零，只要这个数的阶乘结果的分解质因数上有2和5就够了。（一个数n的阶乘含有多少个质因子x ，这个我们可以用
）
扯的远了，没有必要，这个题目，我们只要遍历就行，同时这里遍历肯定不会超时，2这个因子过两个数肯定有，而5这个因子过5个数又肯定会有。

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int>pii;
#define first fi
#define second se
#define  LL long long
#define fread() freopen("in.txt","r",stdin)
#define fwrite() freopen("out.txt","w",stdout)
#define CLOSE() ios_base::sync_with_stdio(false)const int MAXN = 1e5;
const int MAXM = 1e6;
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;int main(){CLOSE();
//  fread();
//  fwrite();LL a,b;cin>>a>>b;if(a==b) {puts("1");return 0;}LL ans=(a+1)%10;for(LL i = a+2;i<=b;i++) {ans=(ans%10*i%10)%10;if(ans==0) break;}printf("%lld\n",ans);return 0;
}

C The Intriguing Obsession
— This is not playing but duty as allies of justice, Nii-chan!

— Not allies but justice itself, Onii-chan!

With hands joined, go everywhere at a speed faster than our thoughts! This time, the Fire Sisters — Karen and Tsukihi — is heading for somewhere they’ve never reached — water-surrounded islands!

There are three clusters of islands, conveniently coloured red, blue and purple. The clusters consist of a, b and c distinct islands respectively.

Bridges have been built between some (possibly all or none) of the islands. A bridge bidirectionally connects two different islands and has length 1. For any two islands of the same colour, either they shouldn’t be reached from each other through bridges, or the shortest distance between them is at least 3, apparently in order to prevent oddities from spreading quickly inside a cluster.

The Fire Sisters are ready for the unknown, but they’d also like to test your courage. And you’re here to figure out the number of different ways to build all bridges under the constraints, and give the answer modulo 998 244 353. Two ways are considered different if a pair of islands exist, such that there’s a bridge between them in one of them, but not in the other.

Input
The first and only line of input contains three space-separated integers a, b and c (1 ≤ a, b, c ≤ 5 000) — the number of islands in the red, blue and purple clusters, respectively.

Output
Output one line containing an integer — the number of different ways to build bridges, modulo 998 244 353.

Example
Input
1 1 1
Output
8
Input
1 2 2
Output
63
Input
1 3 5
Output
3264
Input
6 2 9
Output
813023575
Note
In the first example, there are 3 bridges that can possibly be built, and no setup of bridges violates the restrictions. Thus the answer is 23 = 8.

In the second example, the upper two structures in the figure below are instances of valid ones, while the lower two are invalid due to the blue and purple clusters, respectively.

哎
按照自己翻译的做这个题，样例都过不去，真是（菜）。。最后Google翻译了一下才明白。看懂题目，应该就不难了。
题意：有三个集合，每个集合中有如干个点，在满足每个点不会连相同颜色的点的条件下，我们能够连边的情况数目。
分析：任意两个集合之间连边的时候是不会影响别的集合，所以我们分开求就行。对于任意两个集合，每个点不能够连相同的颜色，就是相当于一个点只能够和一个点一一对应，从每个集合中取的个数不同，情况数就不同，用组合数。

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int>pii;
#define first fi
#define second se
#define  LL long long
#define fread() freopen("in.txt","r",stdin)
#define fwrite() freopen("out.txt","w",stdout)
#define CLOSE() ios_base::sync_with_stdio(false)const int MAXN = 5000+10;
const int MAXM = 1e6;
const int mod = 998244353 ;
const int inf = 0x3f3f3f3f;LL fac[MAXN]; // 阶乘
LL C[MAXN+100][MAXN+100]; // 组合数
void init(){fac[0]=1;for(int i=1;i<=MAXN;i++) fac[i]=fac[i-1]*i%mod;for(int i=0;i<=MAXN;i++) C[i][0]=1;for(int i=1;i<=MAXN;i++){for(int j=1;j<=i;j++)C[i][j]=(C[i-1][j]+C[i-1][j-1])%mod;}//    for(int i=1;i<=10;i++){//        for(int j=1;j<=i;j++)
//            printf("%d ",C[i][j]);
//        puts("");
//    }
}
LL in[5];
int main(){CLOSE();
//  fread();
//  fwrite();init();cin>>in[1]>>in[2]>>in[3];sort(in+1,in+4);LL ans1=0,ans2=0,ans3=0;for(int i=0;i<=in[1];i++) ans1=ans1%mod+C[in[1]][i]*C[in[2]][i]%mod*fac[i]%mod;for(int i=0;i<=in[1];i++) ans2=ans2%mod+C[in[1]][i]*C[in[3]][i]%mod*fac[i]%mod;for(int i=0;i<=in[2];i++) ans3=ans3%mod+C[in[2]][i]*C[in[3]][i]%mod*fac[i]%mod;printf("%lld\n",ans1*ans2%mod*ans3%mod);return 0;
}

E
E题的题解链接

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