VK Cup 2016 - Round 1 (Div. 2 Edition) A B
链接：戳这里
There will be n problems. The i-th problem has initial score pi and it takes exactly ti minutes to solve it. Problems are sorted by difficulty — it's guaranteed that pi < pi + 1 and ti < ti + 1.
A constant c is given too, representing the speed of loosing points. Then, submitting the i-th problem at time x (x minutes after the start of the contest) gives max(0, pi - c·x) points.
Limak is going to solve problems in order 1, 2, ..., n (sorted increasingly by pi). Radewoosh is going to solve them in order n, n - 1, ..., 1 (sorted decreasingly by pi). Your task is to predict the outcome — print the name of the winner (person who gets more points at the end) or a word "Tie" in case of a tie.
You may assume that the duration of the competition is greater or equal than the sum of all ti. That means both Limak and Radewoosh will accept all n problems.
Input
The first line contains two integers n and c (1 ≤ n ≤ 50, 1 ≤ c ≤ 1000) — the number of problems and the constant representing the speed of loosing points.
The second line contains n integers p1, p2, ..., pn (1 ≤ pi ≤ 1000, pi < pi + 1) — initial scores.
The third line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 1000, ti < ti + 1) where ti denotes the number of minutes one needs to solve the i-th problem.
Output
Print "Limak" (without quotes) if Limak will get more points in total. Print "Radewoosh" (without quotes) if Radewoosh will get more points in total. Print "Tie" (without quotes) if Limak and Radewoosh will get the same total number of points.
Examples
input
3 2
50 85 250
10 15 25
output
Limak
input
3 6
50 85 250
10 15 25
output
Radewoosh
input
8 1
10 20 30 40 50 60 70 80
8 10 58 63 71 72 75 76
output
Tie
Note
In the first sample, there are 3 problems. Limak solves them as follows:
Limak spends 10 minutes on the 1-st problem and he gets 50 - c·10 = 50 - 2·10 = 30 points.
Limak spends 15 minutes on the 2-nd problem so he submits it 10 + 15 = 25 minutes after the start of the contest. For the 2-nd problem he gets 85 - 2·25 = 35 points.
He spends 25 minutes on the 3-rd problem so he submits it 10 + 15 + 25 = 50 minutes after the start. For this problem he gets 250 - 2·50 = 150 points.
So, Limak got 30 + 35 + 150 = 215 points.
Radewoosh solves problem in the reversed order:
Radewoosh solves 3-rd problem after 25 minutes so he gets 250 - 2·25 = 200 points.
He spends 15 minutes on the 2-nd problem so he submits it 25 + 15 = 40 minutes after the start. He gets 85 - 2·40 = 5 points for this problem.
He spends 10 minutes on the 1-st problem so he submits it 25 + 15 + 10 = 50 minutes after the start. He gets max(0, 50 - 2·50) = max(0, - 50) = 0 points.
Radewoosh got 200 + 5 + 0 = 205 points in total. Limak has 215 points so Limak wins.
In the second sample, Limak will get 0 points for each problem and Radewoosh will first solve the hardest problem and he will get 250 - 6·25 = 100 points for that. Radewoosh will get 0 points for other two problems but he is the winner anyway.
In the third sample, Limak will get 2 points for the 1-st problem and 2 points for the 2-nd problem. Radewoosh will get 4 points for the 8-th problem. They won't get points for other problems and thus there is a tie because 2 + 2 = 4.
题意：我感觉这道题意还是看hint吧
思路：模拟这个过程就可以了,不知道要不要开long long ,反正开了不吃亏
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
int n,c;
int p[55],t[55];
int main(){
scanf("%d%d",&n,&c);
for(int i=1;i<=n;i++) scanf("%d",&p[i]);
for(int i=1;i<=n;i++) scanf("%d",&t[i]);
ll sum1=0,sum2=0,now=0;
for(int i=1;i<=n;i++){
now=now+t[i];
sum1+=max((ll)0,p[i]-now*c);
}
now=0;
for(int i=n;i>=1;i--){
now+=t[i];
sum2+=max((ll)0,p[i]-now*c);
}
///cout<<sum1<<" "<<sum2<<endl;
if(sum1==sum2) cout<<"Tie"<<endl;
else if(sum1>sum2) cout<<"Limak"<<endl;
else cout<<"Radewoosh"<<endl;
return 0;
}
Spring is starting and the Winter sleep is over for bears. Limak has just woken up and logged in. All his friends still sleep and thus none of them is online. Some (maybe all) of them will appear online in the next hours, one at a time.
The system displays friends who are online. On the screen there is space to display at most k friends. If there are more than k friends online then the system displays only k best of them — those with biggest ti.
Your task is to handle queries of two types:
"1 id" — Friend id becomes online. It's guaranteed that he wasn't online before.
"2 id" — Check whether friend id is displayed by the system. Print "YES" or "NO" in a separate line.
Are you able to help Limak and answer all queries of the second type?
Input
The first line contains three integers n, k and q (1 ≤ n, q ≤ 150 000, 1 ≤ k ≤ min(6, n)) — the number of friends, the maximum number of displayed online friends and the number of queries, respectively.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 109) where ti describes how good is Limak's relation with the i-th friend.
The i-th of the following q lines contains two integers typei and idi (1 ≤ typei ≤ 2, 1 ≤ idi ≤ n) — the i-th query. If typei = 1 then a friend idi becomes online. If typei = 2 then you should check whether a friend idi is displayed.
It's guaranteed that no two queries of the first type will have the same idi becuase one friend can't become online twice. Also, it's guaranteed that at least one query will be of the second type (typei = 2) so the output won't be empty.
Output
For each query of the second type print one line with the answer — "YES" (without quotes) if the given friend is displayed and "NO" (without quotes) otherwise.
Examples
input
4 2 8
300 950 500 200
1 3
2 4
2 3
1 1
1 2
2 1
2 2
2 3
output
NO
YES
NO
YES
YES
input
6 3 9
50 20 51 17 99 24
1 3
1 4
1 5
1 2
2 4
2 2
1 1
2 4
2 3
output
NO
YES
NO
YES
Note
In the first sample, Limak has 4 friends who all sleep initially. At first, the system displays nobody because nobody is online. There are the following 8 queries:
"1 3" — Friend 3 becomes online.
"2 4" — We should check if friend 4 is displayed. He isn't even online and thus we print "NO".
"2 3" — We should check if friend 3 is displayed. Right now he is the only friend online and the system displays him. We should print "YES".
"1 1" — Friend 1 becomes online. The system now displays both friend 1 and friend 3.
"1 2" — Friend 2 becomes online. There are 3 friends online now but we were given k = 2 so only two friends can be displayed. Limak has worse relation with friend 1 than with other two online friends (t1 < t2, t3) so friend 1 won't be displayed
"2 1" — Print "NO".
"2 2" — Print "YES".
"2 3" — Print "YES".
题意：给出n,k,q分别表示n个网络点,k为当前在线的网络点的最大个数,q表示q个操作
思路：我们需要模拟整个q的过程,增删询问,显然只需要优先队列+map模拟就好了
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
struct node{
int v,id;
node(int v=0,int id=0):v(v),id(id){}
bool operator < (const node &a) const{
return v>a.v;
}
};
priority_queue<node> qu;
map<int,int> M;
int n,k,q;
node a[300010];
int main(){
scanf("%d%d%d",&n,&k,&q);
for(int i=1;i<=n;i++) {
scanf("%d",&a[i].v);
a[i].id=i;
M[a[i].v]=0;
}
while(q--){
int x,y;
scanf("%d%d",&x,&y);
if(x==1) {
qu.push(node(a[y].v,y));
M[a[y].v]=1;
while(qu.size()>k) {
node now=qu.top();
qu.pop();
M[now.v]=0;
}
}
else {
if(M[a[y].v]==0) cout<<"NO"<<endl;
else cout<<"YES"<<endl;
}
}
return 0;
}