codeforces 260 div2 A,B,C

A:水的问题。排序结构。看看是否相同两个数组序列。

B：他们写出来1，2，3，4，的n钍对5余。你会发现和5环节。

假设%4 = 0，输出4，否则输出0.

写一个大数取余就过了。

Fedya studies in a gymnasium. Fedya's maths hometask is to calculate the following expression:

(1n + 2n + 3n + 4n) mod 5

for given value of n. Fedya managed to complete the task. Can you? Note that given number n can be extremely large (e.g. it can exceed any integer type of your programming language).

Input

The single line contains a single integer n (0 ≤ n ≤ 10105). The number doesn't contain any leading zeroes.

Output

Print the value of the expression without leading zeros.

Sample test(s)

input
4

output
4

input
124356983594583453458888889

output
0

Note

Operation x mod y means taking remainder after division x by y.

Note to the first sample:

$\text{[math]}$

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <map>
#include <set>
#define eps 1e-9
///#define M 1000100
///#define LL __int64
#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)

using namespace std;

const int maxn = 1000010;

int main()
{
    string s;
    while(cin >>s)
    {
        int n= s.size();
        int cnt = s[0]-'0';
        for(int i = 1; i < n; i++)
        {
            cnt %= 4;
            cnt = (cnt*10+(s[i]-'0'))%4;
        }
        if(cnt%4 == 0)
            cout<<4<<endl;
        else cout<<0<<endl;
    }
}

C：给你一些数。你取了一个数那么比这个数大1，和小1的数字就会被删掉。

问你最大能取到的数的和。

先依据数字进行哈希，然后线性的dp一遍，dp[i][1] = ma(dp[i-2][0], dp[i-2][1]) + vis[i]*i,dp[i][0] = max(dp[i-1][0], dp[i-1][1]).1代表取这个数。0代表不取。注意数据类型要用long long。

Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it.

Given a sequence a consisting of n integers. The player can make several steps. In a single step he can choose an element of the sequence (let's denote it ak) and delete it, at that all elements equal to ak + 1 and ak - 1 also must be deleted from the sequence. That step brings ak points to the player.

Alex is a perfectionist, so he decided to get as many points as possible. Help him.

Input

The first line contains integer n (1 ≤ n ≤ 105) that shows how many numbers are in Alex's sequence.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105).

Output

Print a single integer — the maximum number of points that Alex can earn.

Sample test(s)

input
2
1 2

output
2

input
3
1 2 3

output
4

input
9
1 2 1 3 2 2 2 2 3

output
10

Note

Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points.

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <map>
#include <set>
#define eps 1e-9
///#define M 1000100
///#define LL __int64
#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?
0:x)

using namespace std;

const int maxn = 1000010;

LL vis[maxn];
LL dp[maxn][2];

int main()
{
    int n;
    while(cin >>n)
    {
        int x;
        memset(vis, 0, sizeof(vis));
        memset(dp, 0, sizeof(dp));
        for(int i = 0; i < n; i++)
        {
            scanf("%d",&x);
            vis[x] ++;
        }
        dp[1][1] = vis[1];
        dp[2][1] = vis[2]*2;
        dp[2][0] = dp[1][1];
        for(int i = 3; i <= maxn-10; i++)
        {
            dp[i][1] = max(dp[i-2][0], dp[i-2][1])+vis[i]*i;
            dp[i][0] = max(dp[i-1][0], dp[i-1][1]);
        }
        cout<<max(dp[maxn-10][0], dp[maxn-10][1])<<endl;
    }
    return 0;
}