最近只想喊666，因为我是真得菜，大晚上到网吧打代码还是很不错的嘛
2 seconds
256 megabytes
standard input
standard output
As technologies develop, manufacturers are making the process of unlocking a phone as user-friendly as possible. To unlock its new phone, Arkady's pet dog Mu-mu has to bark the password once. The phone represents a password as a string of two lowercase English letters.
Mu-mu's enemy Kashtanka wants to unlock Mu-mu's phone to steal some sensible information, but it can only bark n distinct words, each of which can be represented as a string of two lowercase English letters. Kashtanka wants to bark several words (not necessarily distinct) one after another to pronounce a string containing the password as a substring. Tell if it's possible to unlock the phone in this way, or not.
The first line contains two lowercase English letters — the password on the phone.
The second line contains single integer n (1 ≤ n ≤ 100) — the number of words Kashtanka knows.
The next n lines contain two lowercase English letters each, representing the words Kashtanka knows. The words are guaranteed to be distinct.
Print "YES" if Kashtanka can bark several words in a line forming a string containing the password, and "NO" otherwise.
You can print each letter in arbitrary case (upper or lower).
ya
4
ah
oy
to
ha
YES
hp
2
ht
tp
NO
ah
1
ha
YES
In the first example the password is "ya", and Kashtanka can bark "oy" and then "ah", and then "ha" to form the string "oyahha" which contains the password. So, the answer is "YES".
In the second example Kashtanka can't produce a string containing password as a substring. Note that it can bark "ht" and then "tp" producing "http", but it doesn't contain the password "hp" as a substring.
In the third example the string "hahahaha" contains "ah" as a substring.
这个题就是暴力，要不s是p的子串，要不存在两个字符串可以拼成s，因为长度都是2，直接暴力枚举啊
#include <bits/stdc++.h>
using namespace std;
int main()
{
string s;
cin>>s;
int n;
scanf("%d",&n);
string q[];
for (int i=; i<n; i++)
cin>>q[i];
for (int i=; i<n; i++)
{
if(q[i]==s)return *puts("YES");
for (int j=; j<n; j++)
if (q[i][]==s[]&&q[j][]==s[])return *puts("YES");
}
puts("NO");
return ;
}
2 seconds
256 megabytes
standard input
standard output
Have you ever tried to explain to the coordinator, why it is eight hours to the contest and not a single problem has been prepared yet? Misha had. And this time he has a really strong excuse: he faced a space-time paradox! Space and time replaced each other.
The entire universe turned into an enormous clock face with three hands — hour, minute, and second. Time froze, and clocks now show the time h hours, m minutes, s seconds.
Last time Misha talked with the coordinator at t1 o'clock, so now he stands on the number t1 on the clock face. The contest should be ready by t2 o'clock. In the terms of paradox it means that Misha has to go to number t2 somehow. Note that he doesn't have to move forward only: in these circumstances time has no direction.
Clock hands are very long, and Misha cannot get round them. He also cannot step over as it leads to the collapse of space-time. That is, if hour clock points 12 and Misha stands at 11 then he cannot move to 1 along the top arc. He has to follow all the way round the clock center (of course, if there are no other hands on his way).
Given the hands' positions, t1, and t2, find if Misha can prepare the contest on time (or should we say on space?). That is, find if he can move from t1 to t2 by the clock face.
Five integers h, m, s, t1, t2 (1 ≤ h ≤ 12, 0 ≤ m, s ≤ 59, 1 ≤ t1, t2 ≤ 12, t1 ≠ t2).
Misha's position and the target time do not coincide with the position of any hand.
Print "YES" (quotes for clarity), if Misha can prepare the contest on time, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
12 30 45 3 11
NO
12 0 1 12 1
YES
3 47 0 4 9
YES
The three examples are shown on the pictures below from left to right. The starting position of Misha is shown with green, the ending position is shown with pink. Note that the positions of the hands on the pictures are not exact, but are close to the exact and the answer is the same.
#include<bits/stdc++.h>
using namespace std;
int main()
{
double a[];
int t1,t2;
scanf("%lf%lf%lf%d%d",&a[],&a[],&a[],&t1,&t2);
t1*=,t2*=;
a[]+=a[]/.;
a[]+=a[]/.;
a[]*=;
if(a[]>)a[]-=.;
sort(a,a+);
int tt1=,tt2=;
if(a[]<t1&&t1<a[])tt1=;
else if(a[]<t1&&t1<a[])tt1=;
if(a[]<t2&&t2<a[])tt2=;
else if(a[]<t2&&t2<a[])tt2=;
if(tt1==tt2)puts("YES");
else puts("NO");
return ;
}
2 seconds
256 megabytes
standard input
standard output
Snark and Philip are preparing the problemset for the upcoming pre-qualification round for semi-quarter-finals. They have a bank of n problems, and they want to select any non-empty subset of it as a problemset.
k experienced teams are participating in the contest. Some of these teams already know some of the problems. To make the contest interesting for them, each of the teams should know at most half of the selected problems.
Determine if Snark and Philip can make an interesting problemset!
The first line contains two integers n, k (1 ≤ n ≤ 105, 1 ≤ k ≤ 4) — the number of problems and the number of experienced teams.
Each of the next n lines contains k integers, each equal to 0 or 1. The j-th number in the i-th line is 1 if j-th team knows i-th problem and 0 otherwise.
Print "YES" (quotes for clarity), if it is possible to make an interesting problemset, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
5 3
1 0 1
1 1 0
1 0 0
1 0 0
1 0 0
NO
3 2
1 0
1 1
0 1
YES
In the first example you can't make any interesting problemset, because the first team knows all problems.
In the second example you can choose the first and the third problems.
暴力二进制模拟下就好的，这样复杂度是够的
#include <bits/stdc++.h>
using namespace std;
int vis[];
int main()
{
int n,k;
cin>>n>>k;
memset(vis,,sizeof(vis));
for(int i=; i<n; i++)
{
int ans=;
for(int j=; j<k; j++)
{
int a;
cin>>a;
ans=ans*+a;
}
vis[ans]++;
}
for(int i=; i<(<<k); i++)
for(int j=; j<(<<k); j++)
if(vis[i]>&&vis[j]>&&((i&j)==))
{
cout<<"YES"<<endl;
return ;
}
cout<<"NO"<<endl;
return ;
}