Dr. Bruce Banner hates his enemies (like others don't). As we all know, he can barely talk when he turns into the incredible Hulk. That's why he asked you to help him to express his feelings.
Hulk likes the Inception so much, and like that his feelings are complicated. They have n layers. The first layer is hate, second one is love, third one is hate and so on...
For example if n = 1, then his feeling is "I hate it" or if n = 2 it's "I hate that I love it", and if n = 3 it's "I hate that I love that I hate it" and so on.
Please help Dr. Banner.
The only line of the input contains a single integer n (1 ≤ n ≤ 100) — the number of layers of love and hate.
Print Dr.Banner's feeling in one line.
1
I hate it
2
I hate that I love it
3
I hate that I love that I hate it
题意：输出一个i hate，再 i love 一直下去；
思路：模拟就好；
#include<bits/stdc++.h> using namespace std; #define ll __int64 #define esp 1e-10 const int N=1e2+10,M=1e6+10,mod=1e9+7,inf=1e9+10; int main() { int x,y,z,i,t; string a="I hate"; string b="I love"; scanf("%d",&x); for(i=1;i<=x;i++) { if(i!=1) printf(" that "); if(i&1) cout<<a; else cout<<b; if(i==x) printf(" it "); } return 0; }
Peter Parker wants to play a game with Dr. Octopus. The game is about cycles. Cycle is a sequence of vertices, such that first one is connected with the second, second is connected with third and so on, while the last one is connected with the first one again. Cycle may consist of a single isolated vertex.
Initially there are k cycles, i-th of them consisting of exactly vi vertices. Players play alternatively. Peter goes first. On each turn a player must choose a cycle with at least 2 vertices (for example, x vertices) among all available cycles and replace it by two cycles with p and x - p vertices where 1 ≤ p < x is chosen by the player. The player who cannot make a move loses the game (and his life!).
Peter wants to test some configurations of initial cycle sets before he actually plays with Dr. Octopus. Initially he has an empty set. In the i-th test he adds a cycle with ai vertices to the set (this is actually a multiset because it can contain two or more identical cycles). After each test, Peter wants to know that if the players begin the game with the current set of cycles, who wins?
Peter is pretty good at math, but now he asks you to help.
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of tests Peter is about to make.
The second line contains n space separated integers a1, a2, ..., an (1 ≤ ai ≤ 109), i-th of them stands for the number of vertices in the cycle added before the i-th test.
Print the result of all tests in order they are performed. Print 1 if the player who moves first wins or 2 otherwise.
3
1 2 3
2
1
1
5
1 1 5 1 1
2
2
2
2
2
In the first sample test:
In Peter's first test, there's only one cycle with 1 vertex. First player cannot make a move and loses.
In his second test, there's one cycle with 1 vertex and one with 2. No one can make a move on the cycle with 1 vertex. First player can replace the second cycle with two cycles of 1 vertex and second player can't make any move and loses.
In his third test, cycles have 1, 2 and 3 vertices. Like last test, no one can make a move on the first cycle. First player can replace the third cycle with one cycle with size 1 and one with size 2. Now cycles have 1, 1, 2, 2 vertices. Second player's only move is to replace a cycle of size 2 with 2 cycles of size 1. And cycles are 1, 1, 1, 1, 2. First player replaces the last cycle with 2 cycles with size 1 and wins.
In the second sample test:
Having cycles of size 1 is like not having them (because no one can make a move on them).
In Peter's third test: There a cycle of size 5 (others don't matter). First player has two options: replace it with cycles of sizes 1 and 4 or 2 and 3.
- If he replaces it with cycles of sizes 1 and 4: Only second cycle matters. Second player will replace it with 2 cycles of sizes 2. First player's only option to replace one of them with two cycles of size 1. Second player does the same thing with the other cycle. First player can't make any move and loses.
- If he replaces it with cycles of sizes 2 and 3: Second player will replace the cycle of size 3 with two of sizes 1 and 2. Now only cycles with more than one vertex are two cycles of size 2. As shown in previous case, with 2 cycles of size 2 second player wins.
So, either way first player loses.
题意：将每个数分解，每次只能将这个数拆成两份 如3可以拆成 （1，2)(2,1)，将前i个数分解，不能分的人输；
思路：因为无论怎么分解最后都是1，因为后面还计算前面的前缀和，判断奇偶即可；
#include<bits/stdc++.h> using namespace std; #define ll __int64 #define esp 1e-10 const int N=1e5+10,M=1e6+10,mod=1e9+7,inf=1e9+10; int a[N]; int main() { int x,y,z,i,t; scanf("%d",&x); for(i=1;i<=x;i++) { scanf("%d",&y); if(y%2==0) a[i]=a[i-1]+1; else a[i]=a[i-1]; } for(i=1;i<=x;i++) { if(a[i]&1) printf("1 "); else printf("2 "); } return 0; }
Thor is getting used to the Earth. As a gift Loki gave him a smartphone. There are n applications on this phone. Thor is fascinated by this phone. He has only one minor issue: he can't count the number of unread notifications generated by those applications (maybe Loki put a curse on it so he can't).
q events are about to happen (in chronological order). They are of three types:
- Application x generates a notification (this new notification is unread).
- Thor reads all notifications generated so far by application x (he may re-read some notifications).
- Thor reads the first t notifications generated by phone applications (notifications generated in first t events of the first type). It's guaranteed that there were at least t events of the first type before this event. Please note that he doesn't read first t unread notifications, he just reads the very first t notifications generated on his phone and he may re-read some of them in this operation.
Please help Thor and tell him the number of unread notifications after each event. You may assume that initially there are no notifications in the phone.
The first line of input contains two integers n and q (1 ≤ n, q ≤ 300 000) — the number of applications and the number of events to happen.
The next q lines contain the events. The i-th of these lines starts with an integer typei — type of the i-th event. If typei = 1 or typei = 2 then it is followed by an integer xi. Otherwise it is followed by an integer ti (1 ≤ typei ≤ 3, 1 ≤ xi ≤ n, 1 ≤ ti ≤ q).
Print the number of unread notifications after each event.
3 4
1 3
1 1
1 2
2 3
1
2
3
2
4 6
1 2
1 4
1 2
3 3
1 3
1 3
1
2
3
0
1
2
In the first sample:
- Application 3 generates a notification (there is 1 unread notification).
- Application 1 generates a notification (there are 2 unread notifications).
- Application 2 generates a notification (there are 3 unread notifications).
- Thor reads the notification generated by application 3, there are 2 unread notifications left.
In the second sample test:
- Application 2 generates a notification (there is 1 unread notification).
- Application 4 generates a notification (there are 2 unread notifications).
- Application 2 generates a notification (there are 3 unread notifications).
- Thor reads first three notifications and since there are only three of them so far, there will be no unread notification left.
- Application 3 generates a notification (there is 1 unread notification).
- Application 3 generates a notification (there are 2 unread notifications).
题意：一个手机最多n个应用，1 x表示 x的应用产生了一个未读的消息；
2 x表示x的应用被全读完；
3 x表示最开始的x个未读信息被读；
思路：利用queue先进先出得到顺序，利用标记数组得到答案，详见代码；
#include<bits/stdc++.h> using namespace std; #define ll __int64 #define esp 1e-10 const int N=3e5+10,M=1e6+10,mod=1e9+7,inf=1e9+10; int a[N]; int com[N]; int un[N]; int th[N]; queue<int>q; int main() { int x,y,z,i,t; scanf("%d%d",&x,&y); int ans=0,maxx=0; for(i=0;i<y;i++) { int u; scanf("%d%d",&u,&a[i]); if(u==1) un[a[i]]++,ans++,q.push(a[i]); else if(u==2) { ans-=un[a[i]]-com[a[i]]; com[a[i]]=un[a[i]]; } else { if(a[i]>=maxx) { int T=a[i]-maxx; while(!q.empty()&&T>0) { T--; int v=q.front(); q.pop(); th[v]++; if(th[v]>com[v]) { ans-=th[v]-com[v]; com[v]=th[v]; } } maxx=a[i]; } } printf("%d ",ans); } return 0; }