codeforces 260 div2 A,B,C_html/css_WEB-ITnose

A:水题，结构体排序后，看两个数组的是否序列相同。

B：分别写出来1，2，3，4，的n次方对5取余。你会发现和对5取余有一个循环节。如果%4 = 0，输出4，否则输出0.

写一个大数取余就过了。

B. Fedya and Maths

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

output

input

124356983594583453458888889

output

Note

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

Note to the first sample:

#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 namespace std int maxn="1000010;int" main string s while>>s)    {        int n= s.size();        int cnt = s[0]-'0';        for(int i = 1; i   <br> 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。  <p></p>  <p>先根据数字进行哈希，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。</p>  <p></p>  <p class="sycode">   </p>
<p class="sycode">    </p>
<p class="sycode">     </p>
<p class="sycode">      </p>
<p class="sycode">       </p>
<p class="sycode">        </p>
<p class="sycode">         C. Boredom        </p>        <p class="sycode">         </p>
<p class="sycode">          time limit per test         </p> 1 second                <p class="sycode">         </p>
<p class="sycode">          memory limit per test         </p> 256 megabytes                <p class="sycode">         </p>
<p class="sycode">          input         </p> standard input                <p class="sycode">         </p>
<p class="sycode">          output         </p> standard output                      <p class="sycode">        </p>
<p> 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.</p>        <p> 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.</p>        <p> Alex is a perfectionist, so he decided to get as many points as possible. Help him.</p>              <p class="sycode">        </p>
<p class="sycode">         Input        </p>        <p> The first line contains integer n (1?≤?n?≤?105) that shows how many numbers are in Alex's sequence.</p>        <p> The second line contains n integers a1, a2, ..., an (1?≤?ai?≤?105).</p>              <p class="sycode">        </p>
<p class="sycode">         Output        </p>        <p> Print a single integer ? the maximum number of points that Alex can earn.</p>              <p class="sycode">        </p>
<p class="sycode">         Sample test(s)        </p>        <p class="sycode">         </p>
<p class="sycode">          </p>
<p class="sycode">           input          </p>          <pre style="代码" class="precsshei">21 2

output

input

31 2 3

output

input

91 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 namespace std int maxn="1000010;LL" vis dp main n while>>n)    {        int x;        memset(vis, 0, sizeof(vis));        memset(dp, 0, sizeof(dp));        for(int i = 0; i   <br>  <br>  <p></p> </eps></set></map></ctime></stack></cmath></queue></string></stdio.h></iomanip></string.h></stdlib.h></iostream></algorithm>