CodeForces 1204 (#581 div 2)

Portal

A.BowWow and the Timetable

• the meaning of problems

The number gives you a binary number, so that you find less than the number of all the powers of 4

• ideas

The first reaction is binary and quaternary conversion

(In fact, no real conversion QwQ)

Due to the corresponding binary two of a quaternary

So we can get the number of bits in a quaternary

Quaternary equal to the number of bits is less than the power of all of the number 4, the analog power decimal 10

Because there are not equal, it is based on the binary determine what this number is not a power of 4

Because 1 ₂ 100 ₂ 10000 ₂ , 4 in addition to the binary power of a first, followed by an even number 0

It is determined whether an even number with 0, if the number is one less

• Code

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=1e6+5;
 4 char s[maxn];
 5 int main()
 6 {
 7     scanf("%s",s+1);
 8     int len=strlen(s+1),flag=0;
 9     int cnt=(len+1)/2;
10     for(int i=2;i<=len;i++)
11         if(s[i]=='0') flag++;
12     if(flag==len-1 && len&1)
13         cnt--;
14     printf("%d\n",cnt);
15 }

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B.Mislove Has Lost an Array

• the meaning of problems

Array there is only $ 1 or $ X $ $

$ X $ is even and $ x / 2 $ must exist in the array

Given $ l, r $ array least $ L $ different numbers, up to $ $ R & lt different numbers

The minimum and maximum number of array seek

• ideas

There is a minimum of $ 1, 2,4 ... L $ $ $ like number, if the number n is less than $ 1 $ filled with

There is a maximum of $ 2, 4 and so on ... $ $ r $ the number, if it is less than $ r $ with the largest number of filled

• Code

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 int main()
 5 {
 6     int n,l,r;
 7     cin>>n>>l>>r;
 8     ll Min=0;
 9     int cnt,i;
10     for(i=1,cnt=1;i<=l;cnt*=2,i++)
11         Min+=cnt;
12     Min+=(n-l);
13 
14     ll Max=0;
15     for(cnt=1,i=1;i<=min(n,r);cnt*=2,i++)
16         Max+=cnt;
17     cnt/=2;
18     if(r<=n)
19         Max+=1ll*(n-r)*cnt;
20     cout<<Min<<' '<<Max<<endl;
21 }

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D.Kirk and a Binary String 

• the meaning of problems

Give you a $ 01 $ string s, t let you find a string, so that

• interval t s string and string the same length all monotone nondecreasing
• t 0 up to the number of strings

• ideas

For a string, the current position of the two cases have $ 0 $

• The current position is 0

　　The current position is 0, for he was the starting point of a non-monotonically decreasing sequences certainly contribute,

　　If the length of the lower becomes 1, it will decrease in most cases (except where all the back 0 1 011 111 the same length)

　　If the number of changes to reduce the ratio of 1,0 unchanged contrary to the second task

• The current position is 1

　　If the value of 0, for his (it is 1) as a starting point monotonically non-decreasing sequence, the length has no effect

　　For at zero monotonically non-decreasing sequence, the length will be affected

Since want more of 0, it would need as much as possible to 1 to a 0, then how can it have no effect?

He would need to length (it is 1) as a starting point monotonically non-decreasing sequences of greater than at zero monotonically non-decreasing sequence length

Because that affect the overall situation of the longest, let even small changes, and nor will the overall impact!

In other words, if you want to become $ $ $ 0 $ 1, we must ensure that all monotonic non-decreasing interval behind the sub-sequence length must equal the number 1

I.e. looking from the rear forward, when the number is greater than the number of $ 1 $ $ $ 0 and $ 1 can become $ $ $ 0

• Code

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 const int maxn=1e6+4;
 5 char s[maxn];
 6 int a[maxn];
 7 int main()
 8 {
 9     scanf("%s",s+1);
10     int len=strlen(s+1);
11     int cnt=0;
12     for(int i=len;i>=1;i--)
13     {
14         if(s[i]=='0')
15             cnt++;
16         else
17         {
18             if(cnt) --cnt;
19             else s[i]='0';
20         }
21     }
22 
23     printf("%s",s+1);
24 }

View Code