Codeforces Round #685 (Div. 2)
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
- 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 ...
- Codeforces Round #354 (Div. 2) ABCD
Codeforces Round #354 (Div. 2) Problems # Name A Nicholas and Permutation standard input/out ...
- 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 ...
- 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..... ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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
- Codeforces Round #650 (Div. 3) C. Social Distance (前缀和)
题意:有一排座位,要求每人之间隔\(k\)个座位坐,\(1\)代表已做,\(0\)代表空座,问最多能坐几人. 题解:我们分别从前和从后跑个前缀和,将已经有人坐的周围的位置标记,然后遍历求每一段连续的\ ...
- 浅谈Webpack模块打包工具二
Webpack插件机制介绍 Loader专注实现资源模块加载,plugin解决其他自动化工作,比如清除dist目录,拷贝静态文件值输出目录,压缩输出代码等,下面列举几个plugin 的例子. Webp ...
- Docker配置文件deamon.json详解
vim /etc/docker/daemon.json { "authorization-plugins": [], "data-root": "&q ...
- Chapter Zero 0.1.1 计算机硬件五大单元
计算机硬件的五大单元 先从CPU说起,CPU的全称如下: 中央处理器(Central Processing Unit,CPU) CPU作为一个特定功能的芯片,内含微指令集, 主机的功能差异,主要参考C ...
- Kubernets二进制安装(9)之部署主控节点控制器controller-manager
kube-controller-manager运行控制器,它们是处理集群中常规任务的后台线程 Controller Manager就是集群内部的管理控制中心,由负责不同资源的多个Controller构 ...
- tensorflow报错:Attempting to fetch value instead of handling error Internal: failed to get device attribute 13 for device 0: CUDA_ERROR_UNKNOWN:
就是在spyder跑上一篇文章的代码然后就报错: Attempting to fetch value instead of handling error Internal: failed to get ...
- Java之一个整数的二进制中1的个数
这是今年某公司的面试题: 一般思路是:把整数n转换成二进制字符数组,然后一个一个数: private static int helper1(int i) { char[] chs = Integer. ...
- codeforces 875B
B. Sorting the Coins time limit per test 1 second memory limit per test 512 megabytes input standard ...
- Qt开发Activex笔记(二):Qt调用Qt开发的Activex控件
若该文为原创文章,转载请注明原文出处本文章博客地址:https://blog.csdn.net/qq21497936/article/details/113789693 长期持续带来更多项目与技术分享 ...
- App icons generator
App icons generator https://appicon.co/ Drag or select an app icon image (1024x1024) to generate dif ...