待补

A


#include <bits/stdc++.h>
using namespace std; int n; int main()
{
int __;
scanf("%d", &__);
while(__ -- )
{
scanf("%d", &n);
if(n == 1) printf("0\n");
else if(n == 2) printf("1\n");
else if(n == 3) printf("2\n");
else if(n % 2 == 0) printf("2\n");
else printf("3\n");
}
return 0;
}

B


#include <bits/stdc++.h>
using namespace std;
const int N = 100 + 20; int n, q;
char str[N]; int main()
{
int __;
scanf("%d", &__);
while(__ -- )
{
scanf("%d%d", &n, &q);
scanf("%s", str + 1);
while(q -- )
{
int l, r;
scanf("%d%d", &l, &r);
int flag = 0;
for(int i = 1; i < l; ++ i)
if(str[i] == str[l])
flag = 1;
for(int i = r + 1; i <= n; ++ i)
if(str[i] == str[r])
flag = 1;
if(flag) printf("YES\n");
else printf("NO\n");
}
}
return 0;
}

C


#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 10; int n, k;
char a[N], b[N];
int ca[30], cb[30]; int main()
{
int __;
scanf("%d", &__);
while(__ -- )
{
for(int i = 0; i < 26; ++ i)
ca[i] = cb[i] = 0;
scanf("%d%d", &n, &k);
scanf("%s%s", a + 1, b + 1);
for(int i = 1; i <= n; ++ i) ca[a[i] - 'a'] ++;
for(int i = 1; i <= n; ++ i) cb[b[i] - 'a'] ++;
int flag = 0;
for(int i = 0; i < 26; ++ i)
{
if(ca[i] < cb[i]) { flag = 1; break;}
if((ca[i] - cb[i]) % k == 0)
ca[i + 1] = ca[i + 1] + ca[i] - cb[i];
else {flag = 1; break;}
}
if(flag) puts("No");
else puts("Yes");
}
return 0;
}

D


#include <bits/stdc++.h>
using namespace std;
typedef long long LL; LL d, k; int main()
{
int __;
scanf("%d", &__);
while(__ -- )
{
scanf("%lld%lld", &d, &k);
LL x = 0, y = 0, res = 0;
while(x * x + y * y <= d * d)
{
if(x < y) x += k;
else y += k;
res ++;
}
if(res % 2 == 0) puts("Ashish");
else puts("Utkarsh");
}
return 0;
}

E1/E2


#include <bits/stdc++.h>
using namespace std; const int N = 1 << 17;
int n, v[N], from[N], a[N]; int query(int op, int i, int j)
{
if(op == 1) printf("AND");
if(op == 2) printf("XOR");
printf(" %d %d\n", i, j); fflush(stdout);
int x;
scanf("%d", &x);
return x;
} void print()
{
printf("!");
for(int i = 1; i <= n; ++ i) printf(" %d", a[i]);
puts("");
} void f(int x, int y)
{
a[x] = a[y] = query(1, x, y);
a[1] = v[x] ^ a[x];
for(int i = 2; i <= n; ++ i)
a[i] = a[1] ^ v[i];
print();
} void g(int x, int y)
{
int v1 = query(1, 1, x);
int v2 = query(1, 1, y);
for(int i = 0; i < 16; ++ i)
{
int u = 0;
if(i == 0) u = (v2 >> i) & 1;
else u = (v1 >> i) & 1;
a[1] |= u << i;
for(int j = 2; j <= n; ++ j)
a[j] |= (u ^ ((v[j] >> i) & 1)) << i;
}
print();
} int main()
{
scanf("%d", &n);
for(int i = 2; i <= n; ++ i) v[i] = query(2, 1, i);
for(int i = 2; i <= n; ++ i)
if(!v[i]) { f(1, i); return 0; }
for(int i = 2; i <= n; ++ i)
if(!from[v[i]]) from[v[i]] = i;
else { f(from[v[i]], i); return 0; }
g(from[1], from[2]);
return 0;
}

2020.11.23

Codeforces Round #685 (Div. 2)的更多相关文章

  1. Codeforces Round #366 (Div. 2) ABC

    Codeforces Round #366 (Div. 2) A I hate that I love that I hate it水题 #I hate that I love that I hate ...

  2. Codeforces Round #354 (Div. 2) ABCD

    Codeforces Round #354 (Div. 2) Problems     # Name     A Nicholas and Permutation standard input/out ...

  3. Codeforces Round #368 (Div. 2)

    直达–>Codeforces Round #368 (Div. 2) A Brain’s Photos 给你一个NxM的矩阵,一个字母代表一种颜色,如果有”C”,”M”,”Y”三种中任意一种就输 ...

  4. cf之路,1,Codeforces Round #345 (Div. 2)

     cf之路,1,Codeforces Round #345 (Div. 2) ps:昨天第一次参加cf比赛,比赛之前为了熟悉下cf比赛题目的难度.所以做了round#345连试试水的深浅.....   ...

  5. Codeforces Round #279 (Div. 2) ABCDE

    Codeforces Round #279 (Div. 2) 做得我都变绿了! Problems     # Name     A Team Olympiad standard input/outpu ...

  6. Codeforces Round #262 (Div. 2) 1003

    Codeforces Round #262 (Div. 2) 1003 C. Present time limit per test 2 seconds memory limit per test 2 ...

  7. Codeforces Round #262 (Div. 2) 1004

    Codeforces Round #262 (Div. 2) 1004 D. Little Victor and Set time limit per test 1 second memory lim ...

  8. Codeforces Round #371 (Div. 1)

    A: 题目大意: 在一个multiset中要求支持3种操作: 1.增加一个数 2.删去一个数 3.给出一个01序列,问multiset中有多少这样的数,把它的十进制表示中的奇数改成1,偶数改成0后和给 ...

  9. Codeforces Round #268 (Div. 2) ABCD

    CF469 Codeforces Round #268 (Div. 2) http://codeforces.com/contest/469 开学了,时间少,水题就不写题解了,不水的题也不写这么详细了 ...

随机推荐

  1. git忽略规则以及.gitignore文件不生效解决办法

    正文 Git忽略规则: #此为注释 – 内容被 Git 忽略 .sample # 忽略所有 .sample 结尾的文件 !lib.sample # 但 lib.sample 除外 /TODO # 仅仅 ...

  2. C++ part8

    1.volatile关键字 在C++中,对volatile修饰的对象的访问,有编译器优化上的副作用: 不允许被编译器优化,提供特殊地址的稳定访问(只从内存中读取). 有序性,编译器进行优化时,不能把对 ...

  3. Virtualbox 安装centos7虚拟机

    Virtualbox 安装centos7虚拟机 一,下载centos7 下载地址:https://mirrors.tuna.tsinghua.edu.cn/centos/7.9.2009/isos/x ...

  4. 慕课网站 & MOOC website

    慕课网站 & MOOC website MOOC, massive open online course Mooc for everyone ! 国家精品课程 & 在线学习平台 慕课平 ...

  5. H5 CSS 悬浮滚动条

    H5 CSS 悬浮滚动条 refs xgqfrms 2012-2020 www.cnblogs.com 发布文章使用:只允许注册用户才可以访问!

  6. CSS clip-path in action

    CSS clip-path in action <!DOCTYPE html> <html lang="en"> <head> <meta ...

  7. Learning Web Performance with MDN

    Learning Web Performance with MDN Web 性能是客观的衡量标准,是加载时间和运行时的感知用户体验. https://developer.mozilla.org/en- ...

  8. Bastion Host (BH)

    Bastion Host (BH) 堡垒机 堡垒主机是专门设计和构造成承受攻击网络上的专用计算机. 该计算机通常承载单个应用程序,例如代理服务器,并且所有其他服务都将被删除或限制以减少对计算机的威胁. ...

  9. DENIEL SOIBIM:如何保持坚持

    丹尼尔·索比姆作为加州理工高材生,在2005年通过创建投资俱乐部对潜力公司进行天使投资,获得了美国Blue Run高层的重视,并相继担任Blue Run潜力营收专家评估师,2009年成为星盟集团的副总 ...

  10. 09_MySQL数据库的索引机制

    CREATE TABLE t_message( id INT UNSIGNED PRIMARY KEY, content VARCHAR(200) NOT NULL, type ENUM(" ...