# 596 div2 AC solution to a problem

A. Forgetting Things

Attendance problems, there are pit

　　Can be directly determined in each case, note that a = 9, b = 1 in the case.

 1 #include <iostream>
 2 using namespace std;
 3 
 4 int main()
 5 {
 6     int a, b;
 7     cin >> a >> b;
 8     if(a == 9 && b == 1)    cout << a << " 10" << endl;
 9     else if(a > b)   cout << "-1" << endl;
10     else if(a == b)  cout << a << "1 " << b << "2" << endl;
11     else if(a + 1 == b)  cout << a << "9 " << b << "0" << endl;
12     else   cout << "-1" << endl;
13     return 0;
14 }

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B2. TV Subscriptions (Hard Version)

Simple version to violence, nothing to say.

Difficulty edition:

I is calculated from the answers to the 1-d, recorded, and define two pointers L R, L point 1, R point d, the case of using the map to record a program for later, to define the record every answer temp

That the map [a [l]] -; if map [a [L]] == 0 then the a [L] program is not purchased, temp--; and L ++;

Let R ++, if the map [a [R]] == 0, the need to purchase, temp ++; and map [a [R]] ++;

Finally ans = min (ans, temp)

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <map>
 4 using namespace std;
 5 const int N = 2e5 + 10;
 6 
 7 int a[N];
 8 
 9 int main()
10 {
11     int t;
12     cin >> t;
13     while(t--)
14     {
15         map<int, int> m;
16         int  n, k, d;
17         cin >> n >> k >> d;
18         for(int i = 1; i <= n; i++) cin >> a[i];
19         int ans = 0;
20         for(int i = 1; i <= d; i++) 
21         {
22             if(m[a[i]] == 0)    ans++;
23             m[a[i]]++;
24         }
25         int tmp = ans;
26         int l = 1, r = d;
27         while(r < n)
28         {
29             m[a[l]]--;
30             if(m[a[l]] == 0)    tmp--;
31             l++; r++;
32             if(m[a[r]] == 0)    tmp++;
33             m[a[r]]++;
34             ans = min(ans, tmp);
35         }
36         cout << ans << endl;
37     }
38     return 0;
39 }

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C. p-binary

Violence did not have to say

n = (2^x1 + p)  + (2^x2 + p ) + ...

Conversion at n - p * cnt 2 ^ x1 + 2 ^ x2 + 2 ^ x3 =;

Obviously violence, enumeration cnt, starting at 1, the boundaries are (n - p * cnt)> 0, because the right> 0;

Setting x = n - cnt * p; x is calculated with a binary number 1, if the number of 1 <= cnt and x is> = cnt, the solution described is the smallest cnt

As for why,

First: the number of revolutions of any one of a binary, i.e., t = 1011, for example, t = 2 ^ 0 + 2 ^ 1 + 2 ^ 3, one can see that each are different, so if the number 1> cnt, then, is It can not match;

　　　As it will be less because may be incorporated i.e. 2 ^ 2 + 2 ^ 2 ^ 3 = 2

Second: x must be greater than or equal cnt, as is the power of 2, i.e., the right side is a minimum value cnt number 2 ^ 0 = cnt, it must be greater than x cnt.

. 1 #include <cstdio>
 2 #include <the cmath>
 . 3 #include <the iostream>
 . 4  the using  namespace STD;
 . 5  
. 6  int n-, P;
 . 7  
. 8  // calculate x has several. 1 
. 9  int CAL ( int x)
 10  {
 . 11      int RES = 0 ;
 12 is      the while (X)
 13 is      {
 14          IF (X & . 1 ) ++ RES ;
 15          X = >> . 1 ;
 16      }
 . 17     return res;
18 }
19 
20 int main()
21 {
22     cin >> n >> p;
23     int ans = 1;
24     while((n - ans * p) > 0)
25     {
26         int x = n - ans * p;
27         int cnt = cal(x);
28         if(cnt <= ans && x >= ans)  
29         {
30             cout << ans << endl;
31             return 0;
32         }
33         ans++;
34     }
35     cout << "-1" << endl;
36     return 0;
37 }

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