Codeforces Round #439 (Div. 2) A B C
强哉qls,这场div2竟是其出的!!!
A. The Artful Expedient
暴力 ^ ,判断是否出现,有大佬根据亦或的性质推出 Karen 必赢,太强啦23333333333333.
#include <stdio.h>
#include <stdlib.h>
#include <cmath>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <string>
#include <ctype.h>
//******************************************************
#define lrt (rt*2)
#define rrt (rt*2+1)
#define LL long long
#define inf 0x3f3f3f3f
#define pi acos(-1.0)
//***************************************************
#define eps 1e-8
#define inf 0x3f3f3f3f
#define INF 2e18
#define LL long long
#define ULL unsigned long long
#define PI acos(-1.0)
#define pb push_back
#define mk make_pair #define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define SQR(a) ((a)*(a))
#define Unique(a) sort(all(a)),a.erase(unique(all(a)),a.end())
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define min4(a,b,c,d) min(min(a,b),min(c,d))
#define max4(a,b,c,d) max(max(a,b),max(c,d))
#define max5(a,b,c,d,e) max(max3(a,b,c),max(d,e))
#define min5(a,b,c,d,e) min(min3(a,b,c),min(d,e))
#define Iterator(a) __typeof__(a.begin())
#define rIterator(a) __typeof__(a.rbegin())
#define FastRead ios_base::sync_with_stdio(0);cin.tie(0)
#define CasePrint pc('C'); pc('a'); pc('s'); pc('e'); pc(' '); write(qq++,false); pc(':'); pc(' ')
#define vi vector <int>
#define vL vector <LL>
#define For(I,A,B) for(int I = (A); I < (B); ++I)
#define rFor(I,A,B) for(int I = (A); I >= (B); --I)
#define Rep(I,N) For(I,0,N)
using namespace std;
const int maxn=+;
int a[maxn],b[maxn],c[maxn*maxn];
int main()
{
int n;
while(cin>>n)
{
Rep(i,n)
{
cin>>a[i];
c[a[i]]=;
}
Rep(i,n)
{
cin>>b[i];
c[b[i]]=;
}
int ans=;
Rep(i,n)
{
Rep(j,n)
{
if( c[ a[i]^b[j]] ) ans++;
}
}
if(ans%) puts("Koyomi");
else puts("Karen");
}
return ;
}
B - The Eternal Immortality
查看 m!/n! 最后一位,水题,没啥好讲的
#include <stdio.h>
#include <stdlib.h>
#include <cmath>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <string>
#include <ctype.h>
//******************************************************
#define lrt (rt*2)
#define rrt (rt*2+1)
#define LL long long
#define inf 0x3f3f3f3f
#define pi acos(-1.0)
//***************************************************
#define eps 1e-8
#define inf 0x3f3f3f3f
#define INF 2e18
#define LL long long
#define ULL unsigned long long
#define PI acos(-1.0)
#define pb push_back
#define mk make_pair #define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define SQR(a) ((a)*(a))
#define Unique(a) sort(all(a)),a.erase(unique(all(a)),a.end())
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define min4(a,b,c,d) min(min(a,b),min(c,d))
#define max4(a,b,c,d) max(max(a,b),max(c,d))
#define max5(a,b,c,d,e) max(max3(a,b,c),max(d,e))
#define min5(a,b,c,d,e) min(min3(a,b,c),min(d,e))
#define Iterator(a) __typeof__(a.begin())
#define rIterator(a) __typeof__(a.rbegin())
#define FastRead ios_base::sync_with_stdio(0);cin.tie(0)
#define CasePrint pc('C'); pc('a'); pc('s'); pc('e'); pc(' '); write(qq++,false); pc(':'); pc(' ')
#define vi vector <int>
#define vL vector <LL>
#define For(I,A,B) for(int I = (A); I < (B); ++I)
#define rFor(I,A,B) for(int I = (A); I >= (B); --I)
#define Rep(I,N) For(I,0,N)
#define REP(I,N) For(I,1,N+1)
using namespace std;
const int maxn=+;
int main()
{
LL a,b;
while(cin>>a>>b)
{
LL sum=;
for(LL i=a+;i<=b;i++)
{
if(i%==)
{
sum=;
break;
}
else
sum=sum*i%;
}
cout<<sum<<endl;
}
return ;
}
C - The Intriguing Obsession
一道组合数学题,问了初三认识的某数学网友得到的思路23333,用记忆化搜索写的
#include <stdio.h>
#include <stdlib.h>
#include <cmath>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <string>
#include <ctype.h>
//******************************************************
#define lrt (rt*2)
#define rrt (rt*2+1)
#define LL long long
#define inf 0x3f3f3f3f
#define pi acos(-1.0)
//***************************************************
#define eps 1e-8
#define inf 0x3f3f3f3f
#define INF 2e18
#define LL long long
#define ULL unsigned long long
#define PI acos(-1.0)
#define pb push_back
#define mk make_pair #define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define SQR(a) ((a)*(a))
#define Unique(a) sort(all(a)),a.erase(unique(all(a)),a.end())
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define min4(a,b,c,d) min(min(a,b),min(c,d))
#define max4(a,b,c,d) max(max(a,b),max(c,d))
#define max5(a,b,c,d,e) max(max3(a,b,c),max(d,e))
#define min5(a,b,c,d,e) min(min3(a,b,c),min(d,e))
#define Iterator(a) __typeof__(a.begin())
#define rIterator(a) __typeof__(a.rbegin())
#define FastRead ios_base::sync_with_stdio(0);cin.tie(0)
#define CasePrint pc('C'); pc('a'); pc('s'); pc('e'); pc(' '); write(qq++,false); pc(':'); pc(' ')
#define vi vector <int>
#define vL vector <LL>
#define For(I,A,B) for(int I = (A); I < (B); ++I)
#define rFor(I,A,B) for(int I = (A); I >= (B); --I)
#define Rep(I,N) For(I,0,N)
#define REP(I,N) For(I,1,N+1)
using namespace std;
const int maxn=+;
#define mod 998244353
LL dp[maxn][maxn];
int dir[][]= { {,},{-,},{,},{,-} }; int solve(int x,int y)
{
if(x==||y==)
return ;
if (dp[x][y] == )
{
dp[x][y]=(y*1ll*solve(x-,y-)+solve(x-,y))%mod;
} return dp[x][y];
} int main()
{
int a,b,c;
while(cin>>a>>b>>c)
{
printf("%I64d\n",solve(a,b)*1ll*solve(b,c)%mod*solve(a,c)%mod);
} return ;
}
Codeforces Round #439 (Div. 2) A B C的更多相关文章
- Codeforces Round #439 (Div. 2)【A、B、C、E】
Codeforces Round #439 (Div. 2) codeforces 869 A. The Artful Expedient 看不透( #include<cstdio> in ...
- Codeforces Round #439 (Div. 2) 题解
题目链接 Round 439 div2 就做了两道题TAT 开场看C题就不会 然后想了好久才想到. 三种颜色挑出两种算方案数其实是独立的,于是就可以乘起来了. E题想了一会有了思路,然后YY出了一种 ...
- Codeforces Round #439 (Div. 2) Problem E (Codeforces 869E) - 暴力 - 随机化 - 二维树状数组 - 差分
Adieu l'ami. Koyomi is helping Oshino, an acquaintance of his, to take care of an open space around ...
- Codeforces Round #439 (Div. 2) Problem C (Codeforces 869C) - 组合数学
— This is not playing but duty as allies of justice, Nii-chan! — Not allies but justice itself, Onii ...
- Codeforces Round #439 (Div. 2) Problem B (Codeforces 869B)
Even if the world is full of counterfeits, I still regard it as wonderful. Pile up herbs and incense ...
- Codeforces Round #439 (Div. 2) Problem A (Codeforces 869A) - 暴力
Rock... Paper! After Karen have found the deterministic winning (losing?) strategy for rock-paper-sc ...
- Codeforces Round #439 (Div. 2)
A. The Artful Expedient 题目链接:http://codeforces.com/contest/869/problem/A 题目意思:给你两个数列,各包含n个数,现在让你从上下两 ...
- 「日常训练」The Intriguing Obsession(CodeForces Round #439 Div.2 C)
2018年11月30日更新,补充了一些思考. 题意(CodeForces 869C) 三堆点,每堆一种颜色:连接的要求是同色不能相邻或距离必须至少3.问对整个图有几种连接方法,对一个数取模. 解析 要 ...
- Codeforces Round #439 (Div. 2) E. The Untended Antiquity
E. The Untended Antiquity 题目链接http://codeforces.com/contest/869/problem/E 解题心得: 1.1,x1,y1,x2,y2 以(x1 ...
- Codeforces Round #439 (Div. 2) C. The Intriguing Obsession
C. The Intriguing Obsession 题目链接http://codeforces.com/contest/869/problem/C 解题心得: 1.由于题目中限制了两个相同 ...
随机推荐
- 如何配置Notepad++的C_C++语言开发环境
相信很多人用notepad++,但把其配置成为C/C++还是需要小折腾一下的.本人在网上找了很长时间,也没有一个统一的答案,而且很多人说的方法根本不管用,而且也不够通用,所以还是自己摸索了一下,分享给 ...
- 构建openssl debug版
一.简介 作为一种安全协议,openssl囊括了主要的密码算法.常用的密钥和证书封装管理功能以及SSL协议,并提供了丰富的应用程序供测试或其它目的使用. 参考: http://www.linuxidc ...
- x-www-form-urlencoded
就是application/x-www-from-urlencoded,会将表单内的数据转换为键值对,比如,name=java&age = 23 postman: 2.ajax传值
- postman模拟登录接口
https://blog.csdn.net/qq_22219911/article/details/80235272
- 633. Sum of Square Numbers
static int wing=[]() { std::ios::sync_with_stdio(false); cin.tie(NULL); ; }(); class Solution { publ ...
- 2018.08.11 洛谷P3224 [HNOI2012]永无乡(线段树合并)
传送门 给出n个带点权的点,支持连边和查询连通块第k大. 这个貌似就是一道线段树合并的裸板啊... 代码: #include<bits/stdc++.h> #define N 100005 ...
- Django入门与实践-第20章:QuerySets(查询结果集)(完结)
http://127.0.0.1:8000/boards/1/ #boards/models.py from django.utils.text import Truncator class Topi ...
- 201709012工作日记--activity与service的通信机制
service生命周期 Service主要包含本地类和远程类. Service不是Thread,Service 是android的一种机制,当它运行的时候如果是Local Service,那么对应的 ...
- (最小生成树)Jungle Roads -- HDU --1301
链接: http://acm.hdu.edu.cn/showproblem.php?pid=1301 http://acm.hust.edu.cn/vjudge/contest/view.action ...
- Amazon成本和产出的衡量方式
Amazon用一种T-Shirt Size 估计的方式来做项目. 产品经理会对每一条需求评估上业务影响力的尺寸,如:XXXL 影响一千万人以上或是可以占到上亿美金的市场,XXL,影响百万用户或是占了千 ...