codeforces260div2A,B,C_html/css

来源：纷纭教育

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

1243569835945834534588888

output

Note

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

Note to the first sample:

#include #include #include #include #include #include #include #include #include #include #include #include #include #define eps 1e-9///#define M 1000100///#define LL __int#define LL long long///#define INF 0x7ffffff#define INF 0x3f3f3f3f#define PI 3.14159265358#define zero(x) ((fabs(x)>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< 
 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。
  
 
 
 
 
 
 C. Boredom 
  
 time limit per test 
 1 second 
  
 memory limit per test 
 256 megabytes 
  
 input 
 standard input 
  
 output 
 standard output 
 
  
 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 
 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 #include #include #include #include #include #include #include #include #include #include #include #include #define eps 1e-9///#define M 1000100///#define LL __int#define LL long long///#define INF 0x7ffffff#define INF 0x3f3f3f3f#define PI 3.14159265358#define zero(x) ((fabs(x)>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<

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