CoderForces Round60-(1117A,1117B,1117C题解)

You are given array

If there are many such subsegments find the longest one.

Input

The first line contains single integer

The second line contains

Output

Print the single integer — the length of the longest subsegment with maximum possible arithmetic mean.

Example

input

Copy

5
6 1 6 6 0

output

Copy

The subsegment

题解：题目意思就是让你找一个序列的子序列平均值最大的（如果均值相同，则找最长的）

这题是水题，其实就是找到最大的那个数Maxnum，然后统计Maxnum连续出现的最多次数；

参考代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn=1e5+10;
 5 int n,a[maxn];
 6 int main()
 7 {
 8     scanf("%d",&n);
 9     int Max=-1,ans=-1,cnt=0;
10     for(int i=1;i<=n;++i) scanf("%d",a+i),Max=max(Max,a[i]);
11     for(int i=1;i<=n;++i)
12     {
13         if(a[i]==Max) cnt++;
14         else ans=max(ans,cnt),cnt=0;
15     }
16     ans=max(ans,cnt); 
17     cout<<ans<<endl;
18     
19     return 0;
20 }

View Code

There are

You have time to use some emotes only

Note that two emotes

You have to make your opponent as happy as possible. Find the maximum possible opponent's happiness.

Input

The first line of the input contains three integers

The second line of the input contains

Output

Print one integer — the maximum opponent's happiness if you use emotes in a way satisfying the problem statement.

Examples

input

Copy

6 9 2
1 3 3 7 4 2

output

Copy

input

Copy

3 1000000000 1
1000000000 987654321 1000000000

output

Copy

1000000000000000000

In the first example you may use emotes in the following sequence:

题解：

意思就是给你n个数，然后现在要从这n个数里面挑选m个，每个数可以多次挑选，但不能连续挑选超过k次，（不同位置的数大小相同也是不同的数）

这题就是找到最大和次大的那个数和最大的数出现的次数，然后判断 k==1？等于一的话，就是最大的数*m, 不等同于1的话，如果最大的数出现的次数

大于1的话，答案也是最大的数*m，如果等于1的话，答案就是k个最大的数，1个次大的数，交替,就是ans=(m-m/(k+1))*Max1+m/(k+1)*Max2;

参考代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 ll n,m,k,val,ans;
 5 int main()
 6 {
 7     cin>>n>>m>>k;
 8     ll Max1=0,Max2=0,cnt=1;
 9     for(int i=1;i<=n;++i)
10     {
11         cin>>val;
12         if(val>Max1) Max2=Max1,Max1=val,cnt=1;
13         else if(val>Max2) Max2=val;
14         else if(val==Max1) cnt++;
15     }
16     if(k==1)
17     {
18         if(cnt==1) ans=(m-m/2)*Max1+(m/2)*Max2;
19         else if(cnt>1) ans=m*Max1; 
20     } 
21     else
22     {
23         if(cnt>1) ans=m*Max1;
24         else if(cnt==1) ans=m/(k+1)*Max2+(m-m/(k+1))*Max1;
25     }
26     cout<<ans<<endl;
27     ///////////////////////////////////
28     return 0;
29 }

View Code

You a captain of a ship. Initially you are standing in a point

You know the weather forecast — the string

Ship coordinates change the following way:

if wind blows the direction U, then the ship moves from
if wind blows the direction D, then the ship moves from
if wind blows the direction L, then the ship moves from
if wind blows the direction R, then the ship moves from

The ship can also either go one of the four directions or stay in place each day. If it goes then it's exactly 1 unit of distance. Transpositions of the ship and the wind add up. If the ship stays in place, then only the direction of wind counts. For example, if wind blows the direction Uand the ship moves the direction L, then from point

You task is to determine the minimal number of days required for the ship to reach the point

Input

The first line contains two integers

The second line contains two integers

It is guaranteed that the initial coordinates and destination point coordinates are different.

The third line contains a single integer

The fourth line contains the string

Output

The only line should contain the minimal number of days required for the ship to reach the point

If it's impossible then print "-1".

Examples

input

Copy

0 0
4 6
3
UUU

output

Copy

input

Copy

0 3
0 0
3
UDD

output

Copy

input

Copy

output

Copy

-1

题解：
　　题目意思就是一艘船要从起点(x1,y1)到达终点(x2,y2),每天的风向是不同的，给出了n天的风向信息（每n天具有周期性），且每天小船可以往一个方向移动一个单位或者不移动；

让你求到达终点的最短时间，如果不能到达就返回-1；

这题由于答案具有单调性，因此可以二分答案m;对于风的速度和船的速度可以分开计算；

我们先计算出在m天风可以使小船移动到的位置，然后其和终点(x2,y2)的曼哈顿距离如果大于天数，则不能，小于等于天数则满足。一直二分答案，求出最小的天数；

 1 #include<bits/stdc++.h>
 2 #define maxl 100010
 3 using namespace std;
 4 typedef long long ll;
 5 #define clr(a,val) memset(a,val,sizeof(a))
 6 #define fi first
 7 #define se second
 8 #define pb push_back
 9 const int INF=0x3f3f3f3f;
10 inline ll read()
11 {
12     ll x=0,f=1;char ch=getchar();
13     while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
14     while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
15     return x*f;
16 }
17 int n;
18 int a[maxl];
19 ll sx,sy,ex,ey,cx=0,cy=0;
20 ll tx[maxl],ty[maxl],sumx[maxl],sumy[maxl];
21 char ch[maxl];
22 ll ans;
23 bool judge(ll mid)
24 {
25     ll fx=mid/n*sumx[n]+sumx[mid%n];
26     ll fy=mid/n*sumy[n]+sumy[mid%n];
27     if(abs(ex-(sx+fx))+abs(ey-(sy+fy))<=mid) return 1;
28     else return 0;
29 }
30 void solve()
31 {
32     ll resx=ex-sx,resy=ey-sy,d=0,res=abs(resx)+abs(resy);
33     for(int i=1;i<=n;i++)
34     {
35         if(tx[i]*resx>0) res--;
36         else res++;
37         if(ty[i]*resy>0) res--;
38         else res++;
39         if(res<=i){ans=i;return;}         
40     }
41     if(resx*cx<=0 && cx!=0) d+=cx;
42     if(resy*cy<=0 && cy!=0) d+=cy;
43     if(d>=n){ans=-1;return;}
44     ll l=1,r=1ll<<62,mid;
45     while(l+1<r)
46     {
47         mid=(l+r)>>1;
48         if(judge(mid)) r=mid;
49         else l=mid;
50     }
51     if(judge(l)) ans=l;
52     else if(judge(r)) ans=r;
53     else ans=-1;
54 }
55 int main()
56 {
57     sx=read();sy=read();ex=read();ey=read();
58     scanf("%d",&n);
59     scanf("%s",ch+1);
60     for(int i=1;i<=n;i++)
61     {
62         if(ch[i]=='U') tx[i]=0,ty[i]=1;
63         if(ch[i]=='D') tx[i]=0,ty[i]=-1;
64         if(ch[i]=='L') tx[i]=-1,ty[i]=0;
65         if(ch[i]=='R') tx[i]=1,ty[i]=0;
66         sumx[i]=sumx[i-1]+tx[i];
67         sumy[i]=sumy[i-1]+ty[i];
68         cx+=tx[i];cy+=ty[i];
69     }
70     solve();
71     cout<<ans<<endl;
72     return 0;
73 }

View Code

CoderForces Round60-(1117A,1117B,1117C题解)

A. Best Subsegment

B. Emotes

题解：

参考代码：

C. Magic Ship