Codeforces Round #345 (Div. 2) 总结

A. Joysticks

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

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

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

Examples

input

3 5

output

6

input

4 4

output

5

Note

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.

题意： 两个电池 初始电量为a1 a2 不充电的话每分钟减少2格电 充电的话每分钟增加1格电 当前只有一根充电线 也就是只能给其中一个电池充电

          当其中任意一个电池电量小于等于0时结束 或者都为1时结束  问 这个过程最多能持续多少分钟

题解： 可以判断 若要这个过程持续 必须使电量大的电池的电量大于等于2 知道这个就很好写了 

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int main(){    int a1,a2;    while(scanf("%d%d",&a1,&a2)!=EOF)    {        int cnt=0;        if(a1>=a2)        {            swap(a1,a2);        }        while(a1<=a2)        {            if(!a1||!a2||(a1<2&&a2<2))                break;            a1++;            a2-=2;            cnt++;            if(a1>=a2)            {                swap(a1,a2);            }        }        printf("%d\n",cnt);    }    return 0;}

B. Beautiful Paintings

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n pictures delivered for the new exhibition. The i-th painting has beauty ai. 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 ai + 1 > ai.

Input

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

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

Output

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

Examples

input

520 30 10 50 40

output

4

input

4200 100 100 200

output

2

Note

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.

题意：给你n个序列，然后让你找到一种排列方式，使得ai<ai+1的数目最多。n<=1000，ai<=1000

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;const int maxm=1e5+10;int vis[maxm];int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        int a;        int sum=0;        memset(vis,0,sizeof(vis));        for(int i=0;i<n;i++)        {            scanf("%d",&a);            vis[a]++;        }        for(int i=0;i<n;i++)        {            int ans=-1;            for(int j=0;j<=1000;j++)            {                if(vis[j])                {                    ans++;                    vis[j]--;                }            }            if(ans>0)            {                sum+=ans;            }            else            {                break;            }        }        printf("%d\n",sum);    }    return 0;}

C. Watchmen

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. There are n watchmen on a plane, the i-th watchman is located at point (xi, yi).

They need to arrange a plan, but there are some difficulties on their way. As you know, Doctor Manhattan considers the distance between watchmen i and j to be |xi - xj| + |yi - yj|. Daniel, as an ordinary person, calculates the distance using the formula .

The success of the operation relies on the number of pairs (i, j) (1 ≤ i < j ≤ n), such that the distance between watchman i and watchmen j calculated by Doctor Manhattan is equal to the distance between them calculated by Daniel. You were asked to compute the number of such pairs.

Input

The first line of the input contains the single integer n (1 ≤ n ≤ 200 000) — the number of watchmen.

Each of the following n lines contains two integers xi and yi (|xi|, |yi| ≤ 109).

Some positions may coincide.

Output

Print the number of pairs of watchmen such that the distance between them calculated by Doctor Manhattan is equal to the distance calculated by Daniel.

Examples

input

31 17 51 5

output

2

input

60 00 10 2-1 10 11 1

output

11

Note

In the first sample, the distance between watchman 1 and watchman 2 is equal to |1 - 7| + |1 - 5| = 10 for Doctor Manhattan and  for Daniel. For pairs (1, 1), (1, 5) and (7, 5), (1, 5) Doctor Manhattan and Daniel will calculate the same distances.

题意：给你n个点，求有多少点对的欧几里得距离=曼哈顿距离

#include<stdio.h>#include<string.h>#include<string>#include<map>#include<algorithm>using namespace std;#define LL __int64map<LL,LL>a;map<LL,LL>b;map<pair<LL,LL>,LL>c;int main(){    LL n;    while(scanf("%I64d",&n)!=EOF)    {        LL sum=0;        LL x,y;        a.clear();        b.clear();        c.clear();        for(LL i=0;i<n;i++)        {            scanf("%I64d%I64d",&x,&y);            sum+=(a[x]+b[y]-c[make_pair(x,y)]);            a[x]++;            b[y]++;            c[make_pair(x,y)]++;        }        printf("%I64d\n",sum);    }    return 0;}