A. Joysticks

Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at a₁ percent and second one is charged at a₂ percent. You can connect charger to a joystick only at the beginning of each minute. In one minute joystick either discharges by 2 percent (if not connected to a charger) or charges by 1 percent (if connected to a charger).

Game continues while both joysticks have a positive charge. Hence, if at the beginning of minute some joystick is charged by 1 percent, it has to be connected to a charger, otherwise the game stops. If some joystick completely discharges (its charge turns to 0), the game also stops.

Determine the maximum number of minutes that game can last. It is prohibited to pause the game, i. e. at each moment both joysticks should be enabled. It is allowed for joystick to be charged by more than 100 percent.

The first line of the input contains two positive integers a₁ and a₂ (1 ≤ a₁, a₂ ≤ 100), the initial charge level of first and second joystick respectively.

Output the only integer, the maximum number of minutes that the game can last. Game continues until some joystick is discharged.

In the first sample game lasts for 6 minute by using the following algorithm:

• at the beginning of the first minute connect first joystick to the charger, by the end of this minute first joystick is at 4%, second is at 3%;
• continue the game without changing charger, by the end of the second minute the first joystick is at 5%, second is at 1%;
• at the beginning of the third minute connect second joystick to the charger, after this minute the first joystick is at 3%, the second one is at 2%;
• continue the game without changing charger, by the end of the fourth minute first joystick is at 1%, second one is at 3%;
• at the beginning of the fifth minute connect first joystick to the charger, after this minute the first joystick is at 2%, the second one is at 1%;
• at the beginning of the sixth minute connect second joystick to the charger, after this minute the first joystick is at 0%, the second one is at 2%.

After that the first joystick is completely discharged and the game is stopped.

B. Beautiful Paintings

There are n pictures delivered for the new exhibition. The i-th painting has beauty a_i. We know that a visitor becomes happy every time he passes from a painting to a more beautiful one.

We are allowed to arranged pictures in any order. What is the maximum possible number of times the visitor may become happy while passing all pictures from first to last? In other words, we are allowed to rearrange elements of a in any order. What is the maximum possible number of indices i (1 ≤ i ≤ n - 1), such that a_i + 1 > a_i.

The first line of the input contains integer n (1 ≤ n ≤ 1000) — the number of painting.

The second line contains the sequence a₁, a₂, ..., a_n (1 ≤ a_i ≤ 1000), where a_i means the beauty of the i-th painting.

Print one integer — the maximum possible number of neighbouring pairs, such that a_i + 1 > a_i, after the optimal rearrangement.

In the first sample, the optimal order is: 10, 20, 30, 40, 50.

In the second sample, the optimal order is: 100, 200, 100, 200.

题目链接：http://codeforces.com/contest/651/problem/B

题意：把a这个数组重新排列，使得序列中 满足条件a_i+1>a_i 的i尽可能的多,输出符合条件的i的数目。

分析：暴力可做，找出对应美丽值相同的最大个数t，n-t即为所求值！

下面给出AC代码：

 #include <bits/stdc++.h>

 using namespace std;

 int a[];

 int main()

 {

     int n,x;

     int ans=,t=-;

     cin>>n;

     for(int i=;i<=n;i++)

     {

         cin>>x;

         a[x]++;

     }

     for(int i=;i<=;i++)

         t=max(t,a[i]);

     cout<<n-t<<endl;

     return ;

 }

3
1 1
7 5
1 5

2

6
0 0
0 1
0 2
-1 1
0 1
1 1

11

 #include <bits/stdc++.h>

 using namespace std;

 typedef long long ll;

 ll n,x,y,ans;

 map<ll,ll> a,b;

 map<pair<ll,ll>,ll> c;

 int main()

 {

     for(cin>>n;cin>>x>>y;)

         ans+=(a[x]++)+(b[y]++)-(c[make_pair(x,y)]++);

     cout<<ans<<endl;

     return ;

 }

 #include <bits/stdc++.h>

 using namespace std;

 struct Node

 {

     int a,b;

 }num[];

 bool cmp1(Node a,Node b)        //第一次排序

 {

     if(a.a==b.a)

     {

         return a.b<b.b;

     }

     return a.a<b.a;

 }

 bool cmp2(Node a,Node b)    //第二次排序

 {

     if(a.b==b.b)

     {

         return a.a<b.a;

     }

     return a.b<b.b;

 }

 int main()

 {

     int n;

     while(~scanf("%d",&n))

     {

         for(int i=;i<n;i++)

         {

             scanf("%d%d",&num[i].a,&num[i].b);

         }

         sort(num,num+n,cmp1);

         __int64 temp=;       //几个是相同的

         __int64 tt=;         //重复的个数

         __int64 res=;

         for(int i=;i<n;i++)

         {

             if(num[i].a==num[i-].a)      //xi == xi-1

             {

                 temp++;

                 if(num[i].b==num[i-].b)

                 {

                     tt++;

                 }

                 else

                 {

                     res-=tt*(tt-)/;   //去重

                     tt=;

                 }

             }

             else

             {

                 res+=temp*(temp-)/;       //排列组合，从temp个两两组合的个数

                 res-=tt*(tt-)/;

                 tt=;

                 temp=;

             }

         }

         if(tt!=)       //判断结尾是不是有些没有去重

         {

             res-=tt*(tt-)/;

         }

         tt=;

         if(temp!=)        //判断结尾有些是不是没有计算

         {

             res+=temp*(temp-)/;

         }

         temp=;

         sort(num,num+n,cmp2);       //第二次排序

         for(int i=;i<n;i++)

         {

             if(num[i].b==num[i-].b)

             {

                 temp++;

             }

             else

             {

                 res+=temp*(temp-)/;

                 temp=;

             }

         }

         if(temp!=)

         {

             res+=temp*(temp-)/;

         }

         printf("%I64d\n",res);

     }

     return ;

 }